Since $f(1^b) = f(1)$ equals $g(1) + h(b)$ for all $b$, we have that $h$ is constant, so that $f$ is constant on powers, in the sense that $b \mapsto f(a^b)$ is constant for every $a$. In particular, $f$ is constant on $\mathbb R_{> 0}$.

I am not sure how to make sense of your proposed equation $f(a^b) = g(a) + h(b)$ for, for example, $a = 0$ and $b = -1$, or $a = -1$ and $b = 1/2$.

This answer is to the original version of the question, where $f$ was assumed to be defined on all of $\mathbb R$, and the equation $f(a^b) = g(a) + h(b)$ was apparently demanded for all $a \in \mathbb R$.

Since $f(1^b) = f(1)$ equals $g(1) + h(b)$ for all $b$, we have that $h$ is constant, so that $f$ is constant on powers, in the sense that $b \mapsto f(a^b)$ is constant for every $a$. In particular, $f$ is constant on $\mathbb R_{> 0}$.

I am not sure how to make sense of your proposed equation $f(a^b) = g(a) + h(b)$ for, for example, $a = 0$ and $b = -1$, or $a = -1$ and $b = 1/2$.