Some preliminary remarks:
- The domain $G$ and the analytic function $g$ are irrelevant. With $w = g(a)$ you are looking for a Möbius transformation $T$ which maps the exterior of the disk $B(-w, r)$ onto the unit disk, with $T(w) = 0$.
- Möbius transformations map disk on the extended complex plane $\hat{\Bbb C} = \Bbb C \cup \{ \infty \}$ to disks or lines on the extended complex plane. $\mathbb C \setminus \overline {B(-w,r)} $ cannot be mapped to the full unit disk, the point $T(\infty)$ will always be missing in the image.
- $T(-w) = 0$ is only possible if $w$ lies in the exterior of $B(-w, r)$, i.e. if $2|w| > r$.
So we can formulate the problem as follows: Given $w \in \Bbb C$ and $r > 0$ with $2|w| > r$, find a Möbius transformation $T$ such that $T \left (\hat{\Bbb C} \setminus \overline {B(-w,r)} \right ) = \Bbb D$ and $T(w) = 0$.
One can start as you did: $T_1(z) = r/(z+w)$ maps $\hat{\Bbb C} \setminus \overline {B(-w,r)} $ onto the unit disk, with $T_1(w) = r/(2w) =: \alpha$. The choose $T_2$ as an automorphism of the unit disk with $T_2(\alpha) = 0$. These automorphism are well-known:
$$
T_2(z) = c\frac{z-\alpha}{1-\overline{\alpha} z}
$$
with an arbitrary factor $c$ of modulus one. Then the composition $T = T_2 \circ T_1$ solves the given problem.
Another option to get the same result is to determine the reflection point $w^*$ of $w$ with respect to the disk $B(-w, r)$. Since Möbius transformations preserve symmetry with respect to a circle or line, $T(w) = 0$ implies $T(w^*) = \infty$. The transformation is therefore of the form
$$
T(z) = d \frac{z-w}{z-w^*}
$$
for some constant $d$ of modulus one. The reflection $w^*$ point of $w$ with respect to a circle $B(z_0, r)$ is determined by the formula
$$
(w^* - z_0)\overline{(w-z_0)} = r^2 \, ,
$$
see for example here. In our case that gives $w^* = -w+ r^2/(2\bar w)$.
