Define $(n\mid x) = \begin{cases} 1 \text{ if } n \text{ divides } x \\ 0 \text{ else } \end{cases}$. Let all variables, if untyped be integers. For every prime $p \in \Bbb{Z}$, we have that the function $f:\Bbb{Z} \to B =\{0,1\}, \ f(x) = \neg(p\mid x)$ is a homomorphism of the ring structure of $\Bbb{Z}$ onto the boolean algebra structure $(\{0,1\}, \vee, \wedge)$.
$$ \neg(p|xy) = \neg(p|x)\wedge \neg (p|y) \\ \neg(p|(x + y)) =\neg (p\mid x) \vee \neg(p \mid y) $$
What's neat is that the boolean algebra $\vee$ is not XOR, it's OR. In terms of XOR ($\oplus$) the second formula would be $\neg (p\mid x) \oplus \neg (p \mid y) \oplus \neg (p \mid x)\wedge \neg (p \mid y)$. So we keep using OR here.
Question. Is every structure-preserving map from $(\Bbb{Z}, +, \cdot, 0,1) \to(B, \vee, \wedge, 0, 1)$ necessarily of the form $f(x) = \neg(p\mid x)$ for some prime $p \in \Bbb{Z}$?
If so, then we could identify the two via a bijection.
Your examples are not homomorphisms, the equality $\neg(p|(x + y)) = \neg(p|x) \vee \neg(p|y)$ is not true. Take $x = p - 1$, $y = 1$, then we have $0$ on the left and $1$ on the right.