In the previous solution the uniqueness question (which was asked) is not addressed.

Of course one cannot conclude that $f=g$, because if $f$ satisfies the equation,
and $p$ is any function such that $p(z)=p(a-z)$ then $fp$ satisfies the equation.
And this is the general description of all solutions because if $f$ and $g$ satisfy the
equation, then their ratio has the property that $p(z)=p(a-z)$.