This is an interesting question, and it depends on the situation where you might decrypt with the "wrong" key.
If two keys $k_1$ and $k_2$ are independently generated, and $c$ is an honestly generated ciphertext under $k_1$, then decrypting $c$ under $k_2$ will result in an error, except with negligible probability. If this weren't the case, it would lead to an attack against AEAD security (the attacker just submits a ciphertext under an independently chosen key). This analysis covers the case of "accidental" or "incidental" decryption under the wrong key.
However, this does not cover the case where $k_1, k_2, c$ are all generated adversarially. (Maybe an attacker shows you $c$ and $k_1$, and since $c$ decrypts successfully under $k_1$ you incorrectly conclude that someone with a different key could not have accepted $c$.) The usual definitions of AEAD don't prevent that. There are natural AEAD schemes (including AES-GCM) where it is possible to generate such $k_1, k_2, c$ such that $c$ decrypts without error under both $k_1$ and $k_2$. This property can indeed cause problems for some applications of AEAD, like password-authenticated key agreement and abuse reporting in encrypted messaging.
If it is hard to come up with any $k_1, k_2, c$ where $c$ decrypts without error under both $k_1$ and $k_2$, then we say that the scheme is key-committing. Sometimes the key-committing property requires providing some additional value (apart from the usual ciphertext and key) to help bind the key to the ciphertext. Key-committing encryption is studied here and here.
