Sketch of a proof - will fill in details but recommend to try and do it:
Step 1: use the Mobius automorphism of the disc $M_1(z)=\frac{z-b_1}{1-\bar b_1 z_1}$, where $b_1^2-(a_1-r_1\bar a_1)b_1-r_1=0, |b_1| <1$ (the quadratic has a root in the unit disc as the absolute value of root product is at most $1-|a_1|<1$) to send the circle $C(a_1,r_1)$ to a circle centered at the origin and leave $\mathbb D$ invariant (this follows from the general formula of the center of a circle under a Mobius transform - in general, it is enough to know this exists and that you can always solve for $b_1$ when $a_1,r_1$ are given as above - note that sending $a_1$ to $0$ with a disc automorphism doesn't guarantee that the image circle has its center at $0$ as only the hyperbolic distance is preserved not the Euclidean one!).
Same with $M_2, z_2$. Then $M_(B(a_1,r_1))=B(0, k_1)$ and same with $M_2$ hence we get $g=M_2 \circ f \circ M_1^{-1}$ conformal map from annulus between $k_1,1$ to the annulus between $k_2,1$ and it is enough to prove that is Mobius
Step 2: It is a standard (proof not that hard using harmonic functions for example and will provide if requested) that this implies $k_1=k_2$ and $g(z)=\alpha z, |\alpha| =1$ or $g(z)=\alpha \frac{k_1}{z}$
(the last part is easy as up to inversion we can assume $g$ maps the small circle to the small circle and the unit circle to the unit circle and then $g/z$ is clearly constant as is non zero and $1$ in absolute value on the boundary, the equality of the radiuses is trickier)
(edit later - sketch of the proof that $A(r,1), A(q,1), r,q <1$ are conformally equivalent iff $r=q$ using harmonic functions - another standard proof can be given using the reflection principle)
Let $f$ conformal equivalence between $A(r,1), A(q,1) ,r,q <1$ and by composing with an inversion if needed we can assume $|f(z)| \to 1, |z| \to 1, |f(z)| \to q, |z| \to r$ and let $u(z)=\log |f(z)|- c\log |z|, c\log r=\log q$ which is harmonic and satisfies $u(z) \to 0, z\to \partial A(1,r)$ hence by the maximum/minimum principle for real harmonic functions $u=0$, $|f(z)|=|z|^c$
But using a local holomorphic logarithm $g_w=\log f$ near any point $w \in A(1,r)$
(may not exist globally on the annulus but exists on a small disc $D_w$) and noting that $|f(z)|^{\frac{1}{c}}=|z|$ we get $|\frac{1}{z}e^{\frac{g_w}{c}}|=1$ on $D_w$ hence $e^{\frac{g_w}{c}}=\alpha_w z, |\alpha_w|=1$ and by diferentiating $\frac{f'}{f}=\frac{c}{z}$ on $D_w$. This doesn't depend on $w$ hence it holds on $A(1,r)$ and by the argument principle (or just integrate on a circle within $A(1,r)$ and use that $f$ is conformal equivalence so LHS=$\pm 2\pi i$, while RHS =$2 \pi i c$) it follows that $\pm 1 =c$ and since $c>0$ as $\log r, \log q <0$, $c=1$ and $r=q$ Done!
A common approach it to find a “conjugate” map which is simpler to investigate. Here we can conjugate $L$ with
$$
T(z) = \frac{z-\alpha}{1-\overline{\alpha}z}
$$
so that the fixed point is transformed to the origin.
If $L$ is a Möbius transformation mapping the unit disk onto itself with $L(\alpha) = \alpha$ then
$$
S = T \circ L \circ T^{-1}
$$
is a Möbius transformation mapping the unit disk onto itself with $S(0) = 0$ and $S'(0) = L'(\alpha)$.
It follows that $S(z) = e^{i\theta} z$, and therefore
$$
L(z) = T^{-1}(e^{i\theta} T(z)) \, .
$$
Best Answer
Some preliminary remarks:
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 $w^* = -w+ r^2/(2\bar w)$.