It looks like there should be a way to do it: $e^{x}$ satisfies $f(x+y)=f(x)f(y)$.
Meanwhile its inverse, the natural logarithm, satisfies a similar looking but inverted equation: $f(xy)=f(x)+f(y)$.
Surely there must be some way to manipulate or derive the second functional equation from the first functional equation, right?
In general you can't, but if you know that $f$ is invertible, then given $a,b$ you can take $x=f^{-1}(a)$, $y=f^{-1}(b)$, and then the first equality gives you $$ f^{-1}(ab)=f^{-1}(f(x)f(y))=f^{-1}(f(x+y))=x+y=f^{-1}(a)+f^{-1}(b). $$ As you can do this for any $a,b$, you have obtained the second equality.