Bitwise XOR of regular languages
You can show that $L$ is regular by using an NFA. Briefly, the NFA is going to simulate DFAs for $A$ and for $B$, guessing the decomposition of the input into $x \oplus y$ as it goes. The same works for any other operation such as AND and OR.

In more detail, suppose that we are given DFAs for $A,B$:


Sets of states $Q^A,Q^B$.
Initial states $q_0^A,q_0^B$.
Final states $F^A,F^B$.
Transition functions $\delta^A,\delta^B$.


We construct an NFA for your language $L$ as follows:


Set of states is $Q^A \times Q^B$.
Initial state is $\langle q_0^A,q_0^B \rangle$.
Final states are $F^A \times F^B$.
Transition function is
$$
\begin{align}
&amp;\delta(\langle q^A,q^B \rangle, 0) = \{\langle \delta^A(q^A,0), \delta^B(q^B,0) \rangle, \langle \delta^A(q^A,1), \delta^B(q^B,1) \rangle \}, \\
&amp;\delta(\langle q^A,q^B \rangle, 1) = \{\langle \delta^A(q^A,0), \delta^B(q^B,1) \rangle, \langle \delta^A(q^A,1), \delta^B(q^B,0) \rangle \}.
\end{align}
$$
You can prove inductively that
$$
\delta(\langle q^A,q^B \rangle, w) = \{ \langle \delta^A(q^A,x), \delta^B(q^B,y) \rangle \mid x \oplus y = w \},
$$
from which the correctness of the construction easily follows.