My goal is to disprove the existence of an isomorphism between $(\mathbb R,+)$ and $(\mathbb R^\*, \times)$.

I proceeded by contradiction. Suppose $f$ is such a map.

Then $$f(0-0)=f(0)f(-0)=-f(0)^2$$
$$ f(0+0)=f(0)^2$$

$$f(0-0)=f(0+0)=f(0)$$

$$-f(0)^2=f(0)^2$$

thus, $$f(0)=0$$

This contradicts the fact that $f(0) \in \mathbb R^\*$

Is this proof right ? It looks overly simple to me and doesn't use the one-to-one property of $f$.

EDIT: the proof is indeed flawed: check answers below.
For those looking for a valid one : let $x=f^{-1}(1)$ and $y=f^{-1}(-1)$. Then $x \neq y$ $$f(2x)=f(x)^2=1^2=(-1)^2=f(2y)$$
Since $f$ is one-to-one, $$2x=2y$$

$$x=y$$