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There are many many such functions (except in the case $a=1,b=0,c

eq 1$). For convenience, let me first assume $a

eq 1$ and consider functions $h:\mathbb{R}\to\mathbb{R}$ which satisfy $$h(ax+b)=ch(x)$$ for all $x\in\mathbb{R}$, not just $x>0$. There is then a unique $d\in\mathbb{R}$ such that $ad+b=d$, namely $d=\frac{b}{1-a}$. Writing $f(y)=h(y+d)$ and substituting $x=y+d$ in the functional equation, we then see that the required functional equation on $h$ is equivalent to $f$ satisfying $$f(ay)=cf(y).$$ This equation has tons of solutions besides just an exponential solution $f(y)=c^{\log_a|y|}$ like you mentioned. Indeed, supposing $a>1$, then for any function $f_0:(-a,-1]\cup[1,a)\to\mathbb{R}$, we can extend to a function $f$ satisfying the functional equation on all of $\mathbb{R}$ by defining $f(y)=c^nf_0(y/a^n)$ for the unique $n\in\mathbb{Z}$ such that $y/a^n\in (-a,-1]\cup[1,a)$ when $y

eq 0$, and $f(0)=0$. (If $c=1$, then the value of $f(0)$ can also be assigned arbitrarily.) When $a>1$ the story is similar, just with $[-1,-a)\cup (a,1]$ instead of $(-a,-1]\cup[1,a)$.

The case $a=1$ similarly has many solutions unless $b=0$: given any function $h_0:[0,b)\to\mathbb{R}$, you can extend to $h$ satisfying the functional equation $h(x+b)=ch(x)$ by defining $h(x)=c^nh_0(x-nb)$ for the unique $n\in\mathbb{Z}$ such that $x-nb\in [0,b)$. If $a=1$ and $b=0$, then the problem is trivial: if $c

eq 1$ then obviously only the zero function works and if $c=1$ then any function works.