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.
Edit. For example, if $b=1$, $a(x)=x$, we obtain $$e^tF(t)=F(t-1).$$ Taking logs, $$t+\phi(t)=\phi(t-1),\quad\phi(t)=\log F(t).$$ Differentiating twice gives that $\phi^{\prime\prime}$ is an (arbitrary) periodic function. The simplest solution is obtained when $\phi^{\prime\prime}={\mathrm{const}}$. Then $\phi(t)=-(1/2)(t^2+t)$, and the solution of your original equation is $$f(x)=\exp\left(-(1/2)(\log^2x+\log x)\right),$$ which satisfies $f(x/e)=xf(x)$.