Conjecture function $g(x)$ is even function?

The conjecture is false.

Consider $f(x) = x$ and $g(x) = 1+x$. Now, for all $x,y \in \mathbb{R}$, $$f(x-y) = x-y$$ and $$f(x)g(y) - f(y)g(x) = x(1+y) - y(1+x) = x + xy - y - xy = x-y,$$ so $f(x-y) = f(x)g(y) - f(y)g(x)$. However, $g$ is not even.