Are there any other functions (not necessarily continuous) satisfying $f(xy)=f(x)+f(y)$ other than $f(x)=A \ln x$ and $f(x)=0$?
After a little thought I came to identify a function $$f : \mathbb{C} \to \mathbb{R}: z \mapsto \arg(z),$$ since $$\arg(z_1z_2)=\arg(z_1)+\arg(z_2).$$ Also $$f: A-\left\{0\right\} \to \mathbb{Z_0^+}$$ where $A$ is set of non-zero polynomials such that $$f(x)=\operatorname{Deg}(\text{polynomial}),$$ since $$\operatorname{Deg}(h(x)g(x))=\operatorname{Deg}(h(x))+\operatorname{Deg}(g(x)).$$
Are there other functions with this property?
If you have a function $f$ such that $f(xy)=f(x)+f(y)$, then $g(x)=f(e^{x})$ satisfies $g(x+y)=g(x)+g(y)$. There are highly discontinuous functions $g$ that satisfy this identity, but they require the Axiom of Choice, which is a standard Axiom of Mathematics. For any such $g$, the function $h(x)=g(\ln x)$ is discontinuous on $(0,\infty)$ and satisfies the following $$ h(xy)=g(\ln(xy))=g(\ln x+\ln y)=g(\ln x)+g(\ln y)= h(x)+h(y),\;\; x,y > 0. $$