If you're not concerned with continuity, integrability, or any of the niceties of real analysis (and there's nothing in the problem that says you are), then, given $a\ne 0$ and $g$, you can let $f$ be any function on the interval $[0,|a|)$ and simply extend it to all real numbers by repeated applications of the given functional equation (written in the form $f(x) = f(x+a) - ag(x)$ to extend it in the direction opposite to the sign of $a$). If $a=0$, it's clear $f$ can be anything.
Addition: Let me be somewhat more explicit. If $a=1$, we can let $f$ be identically zero on $[0,1)$ and equal to $\sum_{k=1}^{[x]} g(x-k)$ for $x\ge1$, with a similar formula for $x<0$.
