Is it safe to distribute different portions of the same data encrypted with different AES keys?
We assume that Bob knows B and Enc(K1, ABC) but doesn't know K1, A or C. Then the only information Bob can learn about ABC is the length of AC (not even the length of A and C individually). If Bob knows Enc(K1, M), Bob can't even tell whether B is a substring of M. This is assuming that Enc is any sensible symmetric cipher, including AES-CBC (with a random IV), AES-CTR (with a initial counter value that doesn't overlap any counter value used for another input with the same key), AES-GCM (with a non-repeated nonce), AES-CCM (with a non-repeated nonce), etc.

AES itself is a permutation on 128-bit blocks, not a way to encrypt a message. involves some xor operations internally, but this is not a xor between the message and something else. There is no simple arithmetic relation involving related inputs and related outputs, otherwise we'd consider it to be completely broken. Given AES_Enc(K, A) and AES_Enc(K, B) where A and B are 128-bit strings, the only thing you can deduce about A and B is whether they're equal or not (A = B if and only if AES_Enc(K, A) = AES_Enc(K, B)).

There are ways to encrypt a message that involve xoring the message with a keystream which is produced using AES as a building block: ciphertext = plaintext ⊕ keystream. AES-CTR is a relatively simple one, where the keystream is built by encrypting successive values. Suppose you have the encryption of two messages with the same key:

ciphertext1 = plaintext1 ⊕ keystream1
ciphertext2 = plaintext2 ⊕ keystream2


keystream1 and keystream2 are indistinguishable from random to someone who doesn't know the key. Therefore ciphertext1 and ciphertext2 are also indistinguishable from random. Any relationship between the plaintexts is completely hidden by this random data, and not visible as a relationship between the ciphertexts.

Badly used modes of operation, for example using a non-random IV, can cause keystream1 and keystream2 to be related in some way. But that's due to badly using the encryption mechanism. The standard mechanisms themselves don't have such weaknesses.

Note that this assumes that Bob doesn't know the key. If Bob knows some part of K1, all bets are off. Never reveal part of a secret key.