For invertible $M \in \mathbb C^{2\times2}$ we can find invertible $S \in \mathbb C^{2\times2}$ so that $SMS^{-1}$ is either diagonal, or triangular with two equal diagonal elements.
There is an isomorphism between $\mathbb C^{2 \times 2} /\mathbb C^+I$ and the group of Mobius Transformations, so the question is:
Prove that invertible matrices can be diagonalised, using the properties of Mobius Transformations
A hint we are given is:
Show that any Mobius Transformation has at least one but at most two fixed points. Identify the matrices with their corresponding Mobius Transformations.
Now clearly the identity fixes more than two points, but it is unique in that respect (since Mobius transforms are uniquely defined by their mapping of three points), and we are also using the convention that the point at infinity is preserved under translations.
I have failed to come up with a construction or argument for the existence of such $S$. I assume it will comprise elementary transformations $z \mapsto az$, $z \mapsto 1/z$, $z \mapsto z + b$.
To be clear we want $S\frac{az + b}{cz + d}S^{-1} = \frac{\alpha z}{\delta}$ or $z + \frac{\beta}{\alpha}$. That is: "Any invertible mobius transformation is, at its core, a scaling or a translation".
Some hints:
The isomorphism between ${\mathbb C}^{2\times2}/{\mathbb C}^*I$ and the group of Moebius transformations has to do with homogeneous coordinates $(z_0,z_1)$ on the $1$-dimensional complex projective space, known as Riemann sphere with its complex coordinate $z$, $z=\infty$ allowed. A point $[z_0,z_1]\in\dot {\mathbb C}^2/{\mathbb C}^*$ corresponds to the point $z={z_0\over z_1}$ on the Riemann sphere, and a linear map with matrix $\left[\matrix{a&b\cr c&d\cr}\right]$ on the $(z_0,z_1)$ space corresponds to the Moebius transformation $T(z)={az+b\over cz+d}$ of the Riemann sphere.