You are asking yourself whether such a map exists at all. The question is: Can a "circle" $\gamma$ be mapped onto itself such that three arbitrarily given points on $\gamma$ are mapped onto another given triple of points on $\gamma$?
Now a familiar example of such a "circle" is the real axis. Given three different points $a<b<c$ in ${\mathbb R}$, the Moebius map
$$S:\quad z\mapsto{z-a\over z-c}$$
maps the real axis (incl. $\infty$) onto itself; furthermore $S(a)=0$, and $S(c)=\infty$. Let $S(b)=:q<0$. Then the map
$$T:\quad z\mapsto {1\over q}\>{z-a\over z-c}\tag{1}$$
maps the arbitrary triple $(a,b,c)$ onto the "special" triple $(0,1,\infty)$. When $(a',b',c')$ is another such triple, and $T'$ the corresponding map $(1)$, then $$T_*:=T'^{-1}\circ T$$ will map $(a,b,c)$ onto $(a',b',c')$.
There is one proviso however: When $(a,b,c)$ and $(a',b',c')$ do not "induce the same orientation" on ${\mathbb R}$ the map $T_*$ will map the upper halfplane onto the lower halfplane. The same is true for your two triples on the unit circle $\partial D$: If they don't induce the same orientation of $\partial D$ the resulting Moebius map $T_*$ does not map $D$ onto itself, but interchanges the interior and the exterior of $D$.
If you think of the domain $D=\mathbb C \setminus {(-\infty,1]\cup[1,\infty)}$, this is not a doubly slit domain on the sphere, but it only has a single slit, passing through infinity (note that $\infty \notin D$). So you will have to move this slit to a slit from $0$ to $\infty$. This can be done with a Mobius transformation that preserves the real line, and maps $1$ to $0$ and $-1$ to $\infty$, say. It is easy to see that $(z-1)/(z+1)$ works.
Now you can unfold $\mathbb C \setminus [0,\infty)$ using $z^{1/2}$ (the branch that maps $-1$ to $i$). The image is the upper half plane, and it is easy now to map it to the unit disk with the Cayley transform $(z-i)/(z+i)$.
Best Answer
With $\frac{z-a}{1-az}$ send $D\setminus(-1,a]$ to $D\setminus(-1,0]$,
with $i z^{1/2}$ send $D\setminus(-1,0]$ to $|z|<1,\Im(z)>0$,
with the inverse of $z\to \frac{z-1}{z+1}$ send $|z|<1,\Im(z)>0$, to $\Re(z)>0,\Im(z)>0$,
with $z^2$ send $\Re(z)>0,\Im(z)>0$ to $\Im(z)>0$,
which is biholomorphic to the unit disk with $\frac{z-i}{z+i}$.
Composing with an automorphism of the unit disk you get $f(i/2) = 0$