$f$ can be any function such that $f(2^k m) = f(2^k)$ for any $k \ge 0$ and odd $m$, where $f(0), f(1), f(2), f(2^2), \dots$ are arbitrary integers.

We label the equation as follows.
$$f(x) + f(y) = f(x + 2xy) + f(y - 2xy) \quad (1)$$
Putting $x = 1$ and $y = n$ in (1), we obtain
$$f(1) + f(n) = f(2n + 1) + f(-n). \quad (2)$$
Putting $x = n$ and $y = -1$ in (1), we obtain
$$f(n) + f(-1) = f(-n) + f(2n - 1). \quad (3)$$
Since $f(1) = f(-1)$, by comparing (2) and (3), we immediately obtain
$$f(2n + 1) = f(2n - 1)$$
for any $n \in \mathbb{Z}$. It follows that $f(m) = f(1)$ for any odd $m$. Using (2), we obtain
$$f(n) = f(-n)$$
for any $n \in \mathbb{Z}$.

Next, we put $x = -2k - 1$ and $y = n$ in (1). Since both $x$ and $x + 2xy = x(1 + 2y)$ are odd, we obtain
$$f(n) = f((4k + 3)n).$$
Replacing $k$ by $-(k+1)$, and using $f(m) = f(-m)$, we get
$$f(n) = f((-4k - 1)n) = f((4k + 1)n).$$
Combining these, we know that $f(n) = f(mn)$ for any odd $m$. Therefore, by writing $n = 2^k m$ where $m$ is odd, we find that
$$f(2^k m) = f(2^k).$$
It remains to check that all such functions satisfy all the conditions. Firstly, $f(-1) = f(1)$ is true. Secondly, Note that $v_2(x + 2xy) = v_2(x(1 + 2y)) = v_2(x)$ and similarly $v_2(y - 2xy) = v_2(y)$. Therefore, we must have
$$f(x + 2xy) = f(x), \quad f(y - 2xy) = f(y).$$
This proves (1), and we are done.