Putting for simplicity $x=e^t, q=e^{-b}, F(t)=f(e^t), A(t)=a(e^t)$, we obtain $$A(t)F(t)=F(t-b).$$ Now you can assign $F$ arbitrarily on any interval of length $b$, for example on $(0,b)$, and this formula defines you a solution everywhere left of this interval. If you want a continuous function, you want $A(b)F(b)=F(0)$, otherwise $F$ is arbitrary on $(0,b)$. Similarly, if you want it smooth etc. If $A(t)\neq 0$ for all $t$, you can also extend your solution to the right.
The answer on further questions depends on what is exactly known about $a$, and what properties you want $f$ to have.