So I have found this thread: $\mathbb{C}\setminus[-1,1]$ is conformally equivalent to $\mathbb{E}\setminus\{0\}$
it says that the function $f(z)=\frac{1}{2}(z+\frac{1}{z})$ from $\mathbb{C} \setminus [-1,1]$ $\rightarrow$ $\mathbb{D}\setminus\{0\}$ has the needed property to map these domains conformally to each other. However I am stuck showing that, so I first need to show that
*) f(z) is injective and surjective, showing that its modulus is always $\leq 1$
*) calculating the inverse function: for the inverse, what I have until now:
$ w = \frac{1}{2}(\frac{z^2+1}{z})$
$ 2wz = z^2+1$ however that would lead to a quadratic equation with the solution $w_{1,2} = z \pm \sqrt{(z-1)(z+1)}$ I do not know whether that is right and how to proceed from here
following the comments, I now know that the approach from the other thread can not lead to the solution, so I would be thankful if someone could help with another approach