Non-exponential functions $f(x)$ satisfying $f(x+c)=\gamma(c)f(x)$ Question:
what can be said about the existence of functions
\begin{align}
f:x\mapsto f(x)&\implies x+c\mapsto \gamma(c)f(x)\\  f(x)\ne f(y)&\iff\frac{f(x)}{\gamma(x)}\ne\frac{f(y)}{\gamma(y)}
\end{align}
These functions would generalize the functional equation $e^{x+y}=e^xe^y$
I am especially interested in the case  $x,f(x)\,\in\,\mathbb{R}$, but also appreciate answers related to different types of functions and arguments.
 A: Assume that $f(x)$ and $g(x)$ are real functions satisfying the relationship $f(a+b)=f(a)g(b)$. Then one must have $f(x)=Kg(x)$ for some constant $K$.
Proof:
Observe that $f(0)=f(0+0)=f(0)g(0)$. So we have that either $f(0)=0$ or $g(0)=1$.
Consider the case where $g(0)=1$. Then we have that $f(a+0)=f(a)g(0)$ but $f(a+0)=f(0+a) = f(0)g(a)$.  So $f(a)g(0)=f(0)g(a)$ and since $g(0)=1$ we have $f(a)=f(0)g(a)$ for all $a$.
Now, consider the case $f(0)=0.$ Then we have $f(x-x)=f(x)g(-x)=0$ for all $x$. But if $f(x)$ is zero for any $x$, then we must have $f(x)$ is always zero. So we must have $g(x)=0$ for all $x$ which also implies that $f(x)=0$.
So our only possible non-trivial cases are $f(x)=Kg(x)$ for some non-zero $K$.
A: Let us consider only the equation $f(x+c)=\gamma(c)f(x)$ for all $x,c$.  Compute
$$
\gamma(a+b)f(x) = f(x+a+b) = \gamma(a)f(x+b) = \gamma(a)\gamma(b)f(x).
$$
Assume there is at least one $x$ with $f(x) \ne 0$.  Then we have, for all $a,b$,
$$
\gamma(a+b) = \gamma(a)\gamma(b) .
\tag{2}$$
Now assume we are in the case $\gamma : \mathbb R \to \mathbb R$.
If $\gamma(a_0) = 0$ for some $a_0$, then $\gamma(a_0 + b) = 0$ for all $b$, and then $\gamma(x) = 0$ for all $x$.  And then $f(x) = 0$ for all $x$.
Assume that $\gamma(x) \ne 0$ for some $x$ so that $\gamma(x) \ne 0$ for all $x$.  From $(2)$ we get $\gamma(x) = \gamma(x/2)^2 > 0$, so $\gamma$ has positive values.  And $\gamma(0) = 1$.
According to the Axiom of Choice, there are badly-behaved solutions of $(2)$.  Let us rule them out by assuming $\gamma$ is differentiable (or, equivalently, continuous, or  Borel measurable, or bounded on some interval, or ...).
Compute
$$
\frac{\gamma(x+h)-\gamma(x)}{h} = \gamma(x)\frac{\gamma(h)-\gamma(0)}{h}
$$
so $\gamma'(x) = k\gamma(x)$ for some constant $k$.  Thus
$$
\gamma(x) = e^{kx}\quad\text{for some constant } k \in \mathbb R .
$$
The OP wants non-exponential functions, so this is not one.

For the case $\gamma : \mathbb C \to \mathbb C$, we get solutions:
$$
\gamma(x) = \exp\left(k x + l \,\overline{x}\right)
\quad\text{for some complex constants }k, l .
$$
in addition to many wild, discontinuous solutions
