That is totally insecure, because you are using raw RSA without padding.

Your application needs a signature, so you should not be dealing with encryptions and decryptions. For instance, PKCS#1 v1.5 is a good protocol, even though the signature is a piece of data that must be appended to what you want to prove the authenticity of.

To verify a PKCS#1 v1.5 signature in Python, you do:

from Crypto.PublicKey import RSA
from Crypto.Signature import PKCS1_v1_5
from Crypto.Hash import SHA

rsa_key = RSA.importKey(open(verification_key_file, "rb").read())
verifier = PKCS1_v1_5.new(rsa_key)
h = SHA.new(data_to_verify)
if verifier.verify(h, signature_received_with_the_data):
    print "OK"
else:
    print "Invalid"

I would strongly recommend to change the PHP code so that it creates such a signature.

There are a few problems already noted in the comments by @Topaco:

Apart from the incorrect key import, the wrong padding is used. The PyCryptodome counterpart to RSA/ECB/PKCS1Padding is PKCS1_v1_5 (and not PKCS1_OAEP). Second, the ciphertext is apparently corrupted: The posted (and thus compromised) private key has a length of 2048 bits = 256 bytes, i.e. the ciphertext must be of the same length. But the posted (Base64 decoded) ciphertext is 380 bytes long (len(by)). Furthermore, in decrypt() not str(by) but by must be passed.

You also have a typo in Pkey, the Base64 encoded body is ...boc61 and not ...boc6 (i.e. the last character is missing). If this is fixed, the key can be imported with RSA.importKey(base64.b64decode(Pkey))

When those are addressed we see that s, after base64 decoding, is too long to be the result of RSA encryption with a 2048 bit modulus. However by trying all offsets into the base64-decoded s and taking the next 256 bytes we get a successful decrypt at offset 3.

# https://stackoverflow.com/q/74840474/238704
import base64

from Cryptodome.Cipher import PKCS1_v1_5
from Cryptodome.PublicKey import RSA
from Cryptodome.Random import get_random_bytes

private_key_pem = '''-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----'''

s = "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"

code_bytes = s.encode('UTF-8')
by = base64.b64decode(code_bytes)
private_key = RSA.import_key(private_key_pem)
cipher = PKCS1_v1_5.new(private_key)
sentinel = get_random_bytes(16)
rsadecrypt = cipher.decrypt(by[3: 3 + 256], sentinel)
if rsadecrypt == sentinel:
    print('failure')
else:
    print(f'success: {rsadecrypt.hex(" ")}')

output is:

success: 48 90 c1 c5 ed fd 67 84 ad 82 df d1 5b 22 40 6f

I don't know what the rest of the bytes of s are all about.

PKCS1 version 1.5 encryption padding is not all that great and is basically deprecated in favor of OAEP padding. One of the weaknesses is the unacceptably high probability that a decryption with the wrong key and/or corrupted ciphertext will succeed. It's unlikely in this case, but not unlikely enough to completely discard the possibility. Although you've provided no additional details about what the payload is supposed to be, the 16 random-looking bytes suggests a key of some sort, perhaps an AES-128 key.

I figured out the solution:

from Crypto.PublicKey import RSA
from Crypto.Signature import PKCS1_v1_5
from base64 import b64decode

rsa_key = RSA.importKey(open('private.txt', "rb").read())
cipher = PKCS1_v1_5.new(rsa_key)
raw_cipher_data = b64decode(<your cipher data>)
phn = cipher.decrypt(raw_cipher_data, <some default>)

This is the most basic form of code. What I learned is first you have to get the RSA_key(private key). For me RSA.importKey took care of everything. Really simple.

Short answer

• the code that you are using doesn't allow you to do that for security reasons
• alternative code below

Long answer

I was curious about your problem and then I started to try to code

After a while I realized that if you run this snippet you will see that it correctly works:

#!/usr/bin/env python

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
import base64

def generate_keys():
    modulus_length = 1024

    key = RSA.generate(modulus_length)
    #print (key.exportKey())

    pub_key = key.publickey()
    #print (pub_key.exportKey())

    return key, pub_key

def encrypt_private_key(a_message, private_key):
    encryptor = PKCS1_OAEP.new(private_key)
    encrypted_msg = encryptor.encrypt(a_message)
    print(encrypted_msg)
    encoded_encrypted_msg = base64.b64encode(encrypted_msg)
    print(encoded_encrypted_msg)
    return encoded_encrypted_msg

def decrypt_public_key(encoded_encrypted_msg, public_key):
    encryptor = PKCS1_OAEP.new(public_key)
    decoded_encrypted_msg = base64.b64decode(encoded_encrypted_msg)
    print(decoded_encrypted_msg)
    decoded_decrypted_msg = encryptor.decrypt(decoded_encrypted_msg)
    print(decoded_decrypted_msg)
    #return decoded_decrypted_msg

def main():
  private, public = generate_keys()
  print (private)
  message = b'Hello world'
  encoded = encrypt_private_key(message, public)
  decrypt_public_key(encoded, private)

if __name__== "__main__":
  main()

but if you now change two of the final lines [i.e. the role of the keys] into:

    encoded = encrypt_private_key(message, private)
    decrypt_public_key(encoded, public)

and rerun the program you will get the TypeError: No private key

Let me quote from this great answer:

"As it turns out, PyCrypto is only trying to prevent you from mistaking one for the other here, OpenSSL or Ruby OpenSSL allow you for example to do both: public_encrypt/public_decrypt and private_encrypt/private_decrypt

[...]

Additional things need to be taken care of to make the result usable in practice. And that's why there is a dedicated signature package in PyCrypto - this effectively does what you described, but also additionally takes care of the things I mentioned"

Adapting this link I came to the following code that should solve your question:

# RSA helper class for pycrypto
# Copyright (c) Dennis Lee
# Date 21 Mar 2017

# Description:
# Python helper class to perform RSA encryption, decryption, 
# signing, verifying signatures & keys generation

# Dependencies Packages:
# pycrypto 

# Documentation:
# https://www.dlitz.net/software/pycrypto/api/2.6/

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
from Crypto.Signature import PKCS1_v1_5
from Crypto.Hash import SHA512, SHA384, SHA256, SHA, MD5
from Crypto import Random
from base64 import b64encode, b64decode
import rsa

hash = "SHA-256"

def newkeys(keysize):
    random_generator = Random.new().read
    key = RSA.generate(keysize, random_generator)
    private, public = key, key.publickey()
    return public, private

def importKey(externKey):
    return RSA.importKey(externKey)

def getpublickey(priv_key):
    return priv_key.publickey()

def encrypt(message, pub_key):
    #RSA encryption protocol according to PKCS#1 OAEP
    cipher = PKCS1_OAEP.new(pub_key)
    return cipher.encrypt(message)

def decrypt(ciphertext, priv_key):
    #RSA encryption protocol according to PKCS#1 OAEP
    cipher = PKCS1_OAEP.new(priv_key)
    return cipher.decrypt(ciphertext)

def sign(message, priv_key, hashAlg="SHA-256"):
    global hash
    hash = hashAlg
    signer = PKCS1_v1_5.new(priv_key)
    if (hash == "SHA-512"):
        digest = SHA512.new()
    elif (hash == "SHA-384"):
        digest = SHA384.new()
    elif (hash == "SHA-256"):
        digest = SHA256.new()
    elif (hash == "SHA-1"):
        digest = SHA.new()
    else:
        digest = MD5.new()
    digest.update(message)
    return signer.sign(digest)

def verify(message, signature, pub_key):
    signer = PKCS1_v1_5.new(pub_key)
    if (hash == "SHA-512"):
        digest = SHA512.new()
    elif (hash == "SHA-384"):
        digest = SHA384.new()
    elif (hash == "SHA-256"):
        digest = SHA256.new()
    elif (hash == "SHA-1"):
        digest = SHA.new()
    else:
        digest = MD5.new()
    digest.update(message)
    return signer.verify(digest, signature)

def main():
    msg1 = b"Hello Tony, I am Jarvis!"
    msg2 = b"Hello Toni, I am Jarvis!"

    keysize = 2048

    (public, private) = rsa.newkeys(keysize)

    # https://docs.python.org/3/library/base64.html
    # encodes the bytes-like object s
    # returns bytes
    encrypted = b64encode(rsa.encrypt(msg1, private))
    # decodes the Base64 encoded bytes-like object or ASCII string s
    # returns the decoded bytes
    decrypted = rsa.decrypt(b64decode(encrypted), private)
    signature = b64encode(rsa.sign(msg1, private, "SHA-512"))

    verify = rsa.verify(msg1, b64decode(signature), public)

    #print(private.exportKey('PEM'))
    #print(public.exportKey('PEM'))
    print("Encrypted: " + encrypted.decode('ascii'))
    print("Decrypted: '%s'" % (decrypted))
    print("Signature: " + signature.decode('ascii'))
    print("Verify: %s" % verify)
    rsa.verify(msg2, b64decode(signature), public)

if __name__== "__main__":
    main()

Final notes:

• the last prints have ascii because as stated here "In case of base64 however, all characters are valid ASCII characters"
• in this case we are using the same key - the private one - both for encrypting and decrypting, so yes: we would end up to be symmetric but...
• but - as stated here - "The public key is PUBLIC - it's something you would readily share and thus would be easily disseminated. There's no added value in that case compared to using a symmetric cipher and a shared key" plus "Conceptually, "encrypting" with the private key is more useful for signing a message whereas the "decryption" using the public key is used for verifying the message"
• the same identical last principle is expressed in this answer - "Typically [...] we say sign with the private key and verify with the public key"