It is easy to understand (even without calculations) that a system of orthogonal circles like in this picture:
transforms into another similar system under a Möbius transformation in the complex plane. So orthogonal angles are conserved for Möbius transformations.
Because the transformation is also continuous, it seems to me that there must be some kind of general argument, that makes it possible to state that every angle is conserved.
Which kind of argument would this be. It probably must involve continuity, but how? A calculation is possible, but I am looking for a general point of view, that uses the conservation of perpendicular angles.
Update (since in my original answer I missed an essential part of the question)
Claim: If ${\bf f}:\>{\mathbb R}^2\to{\mathbb R}^2$ is differentiable at ${\bf 0}$, and $L:=d{\bf f}({\bf 0})$ preserves two disjoint right angles, then $L$ preserves all angles.
Proof. After composing $L$ with a rotation (and a reflection, if necessary) we may assume that $Le_1=\lambda e_1$, $Le_2=\mu e_2$ with $\lambda>0$, $\mu>0$, hence $$[L]=\left[\matrix{\lambda &0\cr 0&\mu \cr}\right]\ .$$ By assumption there are two orthogonal vectors $x:=(a,b)$ and $y:=(-b,a)$ with $ab\ne0$ such that $Lx\perp Ly$. One computes $$Lx=(\lambda a,\mu b),\qquad Ly=(-\lambda b,\mu a)\ .$$ The condition $Lx\perp Ly$ then leads to $$0=Lx\cdot Ly=(-\lambda^2+\mu^2)ab\ ,$$ or $\lambda=\mu$. This shows that $L$ is in fact a similarity.$\qquad\square$
On the other hand there are bijective maps $g:\>S^1\to S^1$ that preserve right angles, but are not just rotations. An example is given by $$g(\phi):=\phi+{\pi\over2}+{1\over 4}\sin^2(2\phi)\ .$$ I leave it to you to verify that this example works.
Original answer:
Here are two ways to go:
(i) Any Moebius transformation $T:\>z\mapsto{az+b\over cz+d}$ can be composed from translations $z\mapsto z+a$, similarities $z\mapsto cz$ and inversions $z\mapsto 1/z$. The first two are obviously conformal, and the conformality of an inversion follows by a simple geometric argument (it is sufficient to assume that one curve is a ray through the origin).
(ii) A Moebius transformation is an analytic function of $z$ (except at $T^{-1}(\infty)$) and therefore locally conformal by general principles about such functions: One has $$T(z_0+Z)-T(z_0)=c Z+o\bigl(|Z|\bigr)\qquad(Z\to0)$$ with $c=T'(z_0)\ne0$. This shows that $dT(z_0)$ is a similarity, hence preserves angles.