It's a very good detailed answer provided by @BruceET. So if you want to do it python, you have to calculate the pooled standard error. I moved the code from this link and you can see it gives you something similar to Bruce's answer:

import numpy as np
from scipy.stats import ttest_ind
from scipy.stats import t
import pandas as pd

def welch_ttest(x1, x2,alternative):
    
    n1 = x1.size
    n2 = x2.size
    m1 = np.mean(x1)
    m2 = np.mean(x2)
    
    v1 = np.var(x1, ddof=1)
    v2 = np.var(x2, ddof=1)
    
    pooled_se = np.sqrt(v1 / n1 + v2 / n2)
    delta = m1-m2
    
    tstat = delta /  pooled_se
    df = (v1 / n1 + v2 / n2)**2 / (v1**2 / (n1**2 * (n1 - 1)) + v2**2 / (n2**2 * (n2 - 1)))
    
    # two side t-test
    p = 2 * t.cdf(-abs(tstat), df)
    
    # upper and lower bounds
    lb = delta - t.ppf(0.975,df)*pooled_se 
    ub = delta + t.ppf(0.975,df)*pooled_se
  
    return pd.DataFrame(np.array([tstat,df,p,delta,lb,ub]).reshape(1,-1),
                         columns=['T statistic','df','pvalue 2 sided','Difference in mean','lb','ub'])

We run this function, i named the lower and upper bounds of the 95% CI as lb and ub.. You can simply modify them in the function:

welch_ttest(ts1,ts2,"equal")


    T statistic     df  pvalue 2 sided  Difference in mean  lb  ub
0   -1.832542   17.90031    0.08356     -1.0    -2.146912   0.146912