It is certainly true that the only $differentiable$ homomorphisms are exponential maps. To see this let $f$ be a differentiable homomorphism, and note that $f(0) = 1$. Write $f'(0) = a$, so
$\lim_{t \to 0} (f(t) - 1)/t = a$. Now we compute $f'(x)$ for arbitrary $x$. We have
$$
f'(x) = \lim_{t \to 0} \frac{f(x+t) - f(x)}{t},
$$
and $f(x+t) = f(x)f(t)$ since $f$ is a homomorphism. This yields
$$
f'(x) = \lim_{t \to 0} f(x)\frac{f(t) - 1}{t} = a f(x).
$$
The unique solution to this differential equation with initial condition $f(0) = 1$ is
$f(x) = \exp(ax)$.