calculus - Differentiable function, not constant, $f(x+y)=f(x)f(y)$, $f'(0)=2$

Let $f: \mathbb R\rightarrow \mathbb R$ a derivable function but not
zero, such that $f'(0) = 2$ and

$$f(x+y)= f(x)\cdot \ f(y)$$ for all
$x$ and $y$ belongs $\mathbb R$. Find $f$.

My first answer is $f(x) = e^{2x}$, and I proved that there are not more functions like $f(x) = a^{bx}$ by Existence-Unity Theorem (ODE), but I don't know if I finished.

What do you think about this sketch of proof's idea?

Thanks,

I'll be asking more things.