The cross-ratio is a Möbius transformation that maps three given distinct points to $0,1,\infty$, respectively. $\DeclareMathOperator{\CR}{CR}$ (Caution: There are different conventions for the cross-ratio, under one convention $\CR(z,z_1,z_2,z_3)$ maps $z_1$ to $0$ and $z_2$ to $1$, under the other widely used one, it maps $z_1\mapsto 1$ and $z_2\mapsto 0$; there may be yet other conventions, but these would be relatively rarely used, I dare say.)
Thus if you have two triples of distinct points, $(z_1,z_2,z_3)$ and $(w_1,w_2,w_3)$, and want to map $z_i\mapsto w_i$, then
$$T = \CR(\,\cdot\,,w_1,w_2,w_3)^{-1}\circ \CR(\,\cdot\,,z_1,z_2,z_3)$$
is the Möbius transformation that achieves that (and that formula is independent on which convention for the cross-ratio you use). Since a Möbius transformation with three fixed points is the identity, $T$ is the only Möbius transformation that maps $z_i\mapsto w_i$ for $i = 1,2,3$.
If you have fewer than $3$ points to map, there is more than one Möbius transformation mapping $z_i\mapsto w_i$ (assuming the $z_i$ resp. $w_i$ are distinct), you can find one Möbius transformation mapping the given points as desired by introducing arbitrary additional points. If you have more than three points to map, there need not exist a Möbius transformation that achieves the desired mapping. Then you pick triples, construct the Möbius transformation mapping these as desired, and check whether the further demands are met.
Generally,
How does one go about proving the existence of a mobius transformation?
is often answered by explicitly giving the desired transformation. Möbius transformations are relatively simple mappings, it is often easy to explicitly produce the desired transformation(s).
One way to do it is to look at specific points on the $x$-axis under $T$.
For example, $T(x) = 0$ for some $x \in \mathbb{R}$, which means that $ax+b=0$ for some $x \in \mathbb{R}$, so there is some $\beta\in\mathbb{R}$ s.t. $a=\beta b$ (here $\beta = -1/x$ so you need to consider that $x$ might be $0$ and handle that separately).
You can go on and show that there are also $\gamma,\delta\in\mathbb{R}$ such that $a=\gamma c, a=\delta d$ and then we get $$T(z) = \frac{az+b}{cz+d} = \frac{az+\beta a}{\gamma az + \delta a} = \frac{z+\beta}{\gamma z + \delta}$$
Best Answer
So you found $|a|=|c|$ and $|b|=|d|$. Note that $f$ remains the same function if we replace $a,b,c,d$ with $\lambda a, \lambda b,\lambda c,\lambda d$ for some $\lambda\ne 0$. We can choose $\lambda$ such that $\lambda b=\overline{\lambda d}$, namely $\lambda=\sqrt{\frac {\overline d}b}$ (whereby incidentally $\lambda\in S^1$). In other words, we may assume wlog that $d=\overline b$. Now consider $\frac {az+b}{cz+\overline b}$ for small real $z$.