A simple example which encodes an array using LabelEncoder, OneHotEncoder, LabelBinarizer is shown below.

I see that OneHotEncoder needs data in integer encoded form first to convert into its respective encoding which is not required in the case of LabelBinarizer.

from numpy import array
from sklearn.preprocessing import LabelEncoder
from sklearn.preprocessing import OneHotEncoder
from sklearn.preprocessing import LabelBinarizer

# define example
data = ['cold', 'cold', 'warm', 'cold', 'hot', 'hot', 'warm', 'cold', 
'warm', 'hot']
values = array(data)
print "Data: ", values
# integer encode
label_encoder = LabelEncoder()
integer_encoded = label_encoder.fit_transform(values)
print "Label Encoder:" ,integer_encoded

# onehot encode
onehot_encoder = OneHotEncoder(sparse=False)
integer_encoded = integer_encoded.reshape(len(integer_encoded), 1)
onehot_encoded = onehot_encoder.fit_transform(integer_encoded)
print "OneHot Encoder:", onehot_encoded

#Binary encode
lb = LabelBinarizer()
print "Label Binarizer:", lb.fit_transform(values)

Another good link which explains the OneHotEncoder is: Explain onehotencoder using python

There may be other valid differences between the two which experts can probably explain.

You just do the transform() on the test data (and do not fit the encoder again). The values that don't occur in "training" datasets would be encoded as 0 in all of the categories (as long as you won't change the handle_unknown parameter). For example:

import category_encoders as ce

train = pd.DataFrame({"var1": ["A", "B", "A", "B", "C"], "var2":["A", "A", "A", "A", "B"]})

encoder = ce.BinaryEncoder(cols = ['var1', 'var2'] , return_df = True)
x_train_data = encoder.fit_transform(train)

#   var1_0  var1_1  var2_0  var2_1
#0  0       1       0       1
#1  1       0       0       1
#2  0       1       0       1
#3  1       0       0       1
#4  1       1       1       0

test = pd.DataFrame({"var1": ["C", "D", "B"], "var2":["A", "C", "F"]})
x_test_data = encoder.transform(test)

#   var1_0  var1_1  var2_0  var2_1
#0  1       1       0       1
#1  0       0       0       0
#2  1       0       0       0

'D' doesn't occur in var1 in training data, so it was encoded as 0 0. 'C' and 'F'don't occur in var2 in training data, so they were both encoded as 0 0.