Suppose your outer circle is the unit circle $C_1=\{z:|z|=1\}$.
All you have to do is to get a linear fractional transformation
taking $C_1$ to itself and the inner circle $C_2$ to a circle centred
at the origin. You might as well rotate $C_2$ so that its centre
is on the real axis. Then $C_2$ meets the real axis in points $a$ and $b$
with $-1 < a < b < 1$. The required linear fractional transformation
will have real coefficients and take $-1$, $a$, $b$, $1$ to $-1$,
$-c$, $c$, $1$ where $0 < c < 1$. The typical linear fractional transformation
fixing $-1$ and $1$ has the form
$$f_u:z\mapsto\frac{z+u}{uz+1}.$$
We are considering $u$ real and for $u$ to take the unit disc to itself,
$|u| < 1$. We need $f_u(a)=-f_u(b)$. This is a quadratic equation in $u$ with two
real roots, one of which satsifies $|u| < 1$.
Well, the relevant line in Ahlfohrs just says "For two-sided arcs the same will be true with obvious modifications. This is the second edition (1966), chapter 6, section 1.3 "Use of the Reflection principle," pages 225-226, just after the Riemann Mapping Theorem.
So, take your domain (you need to type in something about $r e^{i \theta}$ with $0 \leq r < 1$). First, take the principal branch of the square root. That maps your domain to the semicircle $|z| < 1, \; \mbox{Re}\; z > 0.$ Then we have a domain bounded by exactly two free one-sided analytic arcs. So, Theorem 4, section 1.4, page 227, both the boundary line segment and the boundary semicircle are mapped 1-1 and analytically to the boundary circle of the unit disk.
However, this says to me that the original slit does not have a single-valued map to the unit circle, it is evidently double-valued, just as the square root. Maybe it's just me.
EDIT: I have it correct. See pages 18-19 in Boundary Behavior of Conformal Maps by Christian Pommerenke. What he does is switch the order, the conformal map starts in the unit disk, denoted $$f : \mathbb D \rightarrow G.$$ Note: Falcao of Atletico Madrid just scored on Real Madrid, drawn 1-1 at minute 56. OK, the boundary of $\mathbb D$ is the unit circle, denoted $\mathbb T.$
Continuity Theorem. The function $f$ has a continuous extension to $\mathbb D \cup \mathbb T$ if and only if $\partial G$ is locally
connected.
Caratheodory Theorem. The function $f$ has a continuous and injective extension to $\mathbb D \cup \mathbb T$ if and only if
$\partial G$ is a Jordan curve.
If $\partial G$ is locally connected but not a Jordan curve then
parts of $\partial G$ are run through several times. Some of the
possibilities are indicated in Fig. 2.1 where points with the same
letters correspond to each other. Note that e.g. the arcs $a_2 d_1$
and $a_1 d_2$ are both mapped onto the segment $ad.$
Best Answer
Let $D = \mathbb C \setminus \{a\}$. We want to show that $D$ is not conformally equivalent to any annulus. We will achieve this by showing that if a holomorphic map $f: D \rightarrow \mathbb C$ is injective, then its image is unbounded.
Suppose $f: D \rightarrow \mathbb C$ is holomorphic and injective. By the Caserati-Weierstrass Theorem, $a$ is not an essential singularity (because otherwise $f$ would not be injective). Alternatively, you can use Picard's Great Theorem to rule this out.
Case 1. $a$ is a removable singularity. In this case $f$ extends to a holomorphic map $F: \mathbb C \rightarrow \mathbb C$ and has unbounded image by Liouville's Theorem, so also $f$ has unbounded image. So $f$ is not a conformal equivalence between $D$ and an annulus.
Case 2. $a$ is a pole. But then $f$ is unbounded near $a$, so its image is not an annulus.