This is not true without additional hypotesis. If it were true,
it would imply Caratheodory's claim about the existence of a single valued inverse, which is false in general:
Let $f(z)=z^2(z-1)$ on $\mathbb{C}$. This map is surjective, and it has only two singular points: $0$ and $\frac23$. However, if we remove this values from the domain, we get a surjective map from $\mathbb{C}-\{0,\frac23\}$ to $\mathbb{C}$ which is non injective. Using the
fact that $\mathbb{C}-\{0,\frac23\}$ has $\mathbb{D}$ as its universal cover, we get a map
$\varphi:\mathbb{D}\twoheadrightarrow \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and not injective: thus $\varphi$ does not admit a single valued inverse on $G_w^*=Gw=\mathbb{C}$ as such a function should be entire and bounded and thus by Liouville's theorem, constant.
One can also construct $\varphi:\mathbb{C}\twoheadrightarrow \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and non linear (and thus non injective, as the only injective entire functions are linear): see this mathoverflow post for such an example (the error function is an explicit example).
$\varphi$ does not admit a single valued inverse on $G_w^*=G'w=\mathbb{C}$, as such a function should be entire and it should avoid more than one point (since for almost every $w, \varphi^{-1}(w)$ is infinite)
and thus by Picard's little theorem it should be constant.
If, however, we let $f$ satisfy a few more assumptions, the result follows. For example, if $f:G_z\to G_w^*$ is proper, it is a well known result that $f$ is actually a covering map. As $G_w^*$ is simply connected, it's its own universal cover, and thus the holomorphic lifting lemma implies the existence of an inverse and thus in particular the possibility of analytic continuation along polygonal paths.
In general, no. Suppose we take a ${}_2F_1(a,b;c;z)$, continue it analytically, starting at $z=0$, into the lower half-plane, cross the real axis above $1$, then continue in the upper half-plane back to $0$. We could end up with a singularity at $0$. The differential equation for ${}_2F_1(a,b;c;z)$ has singularities at $0$, $1$, and $\infty$. But $\log(z-1)$ has singularities only at $1$ and $\infty$.
Example. The differential equation
$$
4z(1-z)u''(z)+(4-8z)u'(z)-u(z) = 0
\tag1$$
has general solution near $z=0$ given by $C_0 u_0(z)+C_1u_1(z)$,
where
\begin{align}
u_0(z) &= {}_2F_1\left(\frac12,\frac12;1;z\right)
=1+\frac{1}{4}z + \frac{9}{65}z^2+\dots
\\
u_1(z) &= \pi\;{}_2F_1\left(\frac12,\frac12;1;1-z\right) =
\ln\frac{1}{z}+4\ln 2+\dots
\end{align}
Start with the function $u_0$ and analytically continue it along the circle
$$
z = 1-e^{it},\quad 0 \le t < 2\pi .
\tag2$$
When we arrive back at $0$ we are on the branch
$u_0 - \frac{2i}{\pi} u_1$ . This branch goes to $\infty$ as we approach $0$. Here is a graph of the imaginary part of the function along that curve:
The red part is the imaginary part of the image of the lower half of the circle $(2)$, starting at $z=0$; the blue part is the imaginary part
of the image of the upper half of the circle, logarithmically approaching $-\infty$ as $t$ approaches $2\pi$.
Summary: the analytic continuation of the hypergeometric function ${}_2F_1\left(\frac12,\frac12;1;z\right)$, going around the point $z=1$, is a different solution of $(1)$ when we arrive back at $0$.
Best Answer
I think now I can answer my question.
In general, if $\Omega\subset\mathbb C$ is a simply-connected region and $f$ is an analytic function on disc $a\in D\subset \Omega$ such that the germ $[f]_a$ at $a$ of $f$ admits an analytic continuation along any curve $\gamma$ in $\Omega$ from $a$ to $z$. Then, there exists an analytic function $F:\Omega\rightarrow \mathbb C$ with $F{|_D}=f$.
For the proof, let $\gamma$ be a curve in $\Omega$ from $a$ to $z$, and let $\tilde{\gamma}$ be the analytic continuation of $[f]_a$ (i.e the lifting to $\mathcal O$).
Define, $F(a):=\tilde\gamma(1)(z)$ i.e, the value of the germ $\tilde\gamma(1)$ at $z$ (recall that $\tilde\gamma(1)$ is the germ of some function at $z$)
Now, since any two curves $\gamma_1,\gamma_2$ from $a$ to $z$ are homotopic, then $\tilde\gamma_1(1)=\tilde\gamma_2(1)$ and hence the definition of $F$ is well-defined.
For continuity; We note that along any curve $\gamma$ in $\Omega$ and for all $z\in \gamma([0,1])$ we have $F(z)=\tilde\gamma(t)(\gamma(t))$. i.e, $F$ is continuous along all curves in $\Omega$ thus continuous.
It remains to show that $F$ is analytic. Well, at any $z_0\in \Omega$, let the function element $(V,h)$ be a representative of the germ $\tilde \gamma(1)$ where $z_0\in V$. Hence, at a small neighborhood $z_0\in W\subset V$ , $F(z)=h(z)$. (i.e, at each point $z_0$, $F$ admits power series expansion).