I would like to use a measure ${\rm d} \mu (z)$ on ${\mathbb C}$ so that for any $f(z)$ $$\int_{\mathbb C} f(z) {\rm d} \mu (z)$$ is invariant under Möbius transformations.
Taking the transformation as $$M(z) = \frac{a z + b}{cz + d}$$ the measure also might be invariant only under the special Möbius transformations with $ad - bc = 1$. Does such a measure exist?
No. The Mobius transformations include as special cases scalings and translations. Consider now the square $[-1,1) \times [-1,1)$. If $\mu$ is invariant under the Mobius transformations we must have
$$ \mu([-1,1)\times [-1,1)) = \mu([0,1)\times[0,1)) $$
while
$$ \mu([-1,1)\times [-1,1)) = \mu([0,1)\times[0,1)) + \mu([0,1)\times[-1,0)) + \mu([-1,0)\times[0,1)) + \mu([-1,0)\times[-1,0)) = 4 \mu([0,1)\times[0,1)) $$
implying $\mu([-1,1)\times [-1,1)) = 0$. The same argument shows if any open bounded set $A$ is measurable, it would have measure $\mu(A) = 0$.
Thus by sigma-sub-additivity you have that $\mu$ can only be the zero measure if there exists some open $A$ with $\mu(A) < \infty$.
On the other hand, the counting measure is invariant under Mobius transformations.