It can be shown, and is a problem in Rudin's Principles of Mathematical Analysis (Chapter 8), that when $f$ is continuous, and $f(x)f(y) = f(x+y)$, $f$ is a function of the form $e^{cx}$.
Must this necessarily be true when the right hand side is a different function $g$, and then $g = f$? If so, is there a good reason why, in this particular functional equation, it doesn't matter whether it's $f$ or $g$ on each side--but it matters in others?
EDIT: $f$ should also be assumed to be never zero.
For every $x$,
$$ f(x)f(0) = g(x + 0) = g(x) $$
so $g(x) = Cf(x)$, where $C = f(0)$.
Then, you have
$$ f(x)f(y) = Cf(x + y) $$
and further, we can obviously write $f(x) = \dfrac{1}{C}f(x)f(0)$.
Now, if we use the limit definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{\dfrac{1}{C}f(x)f(h) - \dfrac{1}{C}f(x)f(0)}{h} = \dfrac{1}{C}f(x)\dfrac{f(h) - f(0)}{h} = \dfrac{1}{C}f'(0)f(x) $$
So solutions are just $f(x) = Ce^{\alpha x}$; i.e. you get basically the same result, except that you are not forced to take $f(0) = 1$.