The following result is well-known:
Let $U\subset \mathbb{C}$ be simply connected and let $f\colon U \to \mathbb{C}^\ast$ be holomorphic, then there exists $g\colon U \to \mathbb{C}$ holomorphic such that $\operatorname{exp}(g(z))=f(z)$ for all $z\in U$.
I am wondering if (and if so, why) this stays true if we only assume that $f$ is smooth. Feel free to restrict your attention to open balls. This answer suggests that this is true. It would also help, if someone could give the details for why $$ (log(f))(x)=\int_{\gamma} \frac{df}{f} $$ is well-defined for $\gamma$ a differentiable path in a manifold $M$ with vanishing first cohomology group connecting a fixed point $x_0$ and $x$, and why it is smooth with $exp(log(f)(x))=f(x)$ (see this answer).
Sure. The point is that $\exp : \mathbb{C} \to \mathbb{C}^*$ is a covering map. Hence, by the lifting theorem, if $U \subset \mathbb{C}$ is simply connected and $f : U \to \mathbb{C}^*$ is continuous, it can be lifted to $\mathbb{C}$, i.e. there exists $g : U \to \mathbb{C}$ (continuous) such that $f(z) = \exp(g(z))$. Since $\exp$ is analytic, this $g$ will be analytic if $f$ is so, and smooth if $f$ is so.