There is another thread about this topic but I'm interested if the following reasoning is correct:
Let $f:\mathbb C^*\to\mathbb C^*$ be a continuous function that satisfies $f^2(z)=z$. We know that there is a holomorphic square root $g(z)=\exp(\frac{1}{2}\log(z))$ on the slit plane $\mathbb C^-$. Now we can observer that $$\frac{f^2(z)}{g^2(z)}=\frac{z}{z}=1$$ and thus $f^2=g^2$, so either $f=g$ or $f=-g$. The case $f=g$ can't happen because then $g$ could be continuously extended on $\mathbb C^*$ which we know isn't possible. And the case $f=-g$ would mean that $-g$ could be continously extended and therefor $g$ as well?
That proof is incomplete. You did not justify that$$f^2=g^2\implies f=g\vee f=-g.$$Of course, for each individual $z$ it is true that $f(z)=g(z)$ or $f(z)=-g(z)$, but this is not enough to deduce that $f=g$ or $f=-g$.