Is the following statement is true /false ?
Every Möbius transformation $ T_{a,b,c,d}(z) = \frac{(az+b)}{(cz+d)}$ such that $|c|=|d| $ carries the unit circle D onto a straight line.
I think this statement is true because every Möbius transformation is the composition of translation, dilation and inversion.
This appears to be false. If we take $a = 0, b= 1$, and $c = d = 1$, this is the Mobius transformation $T(z) = \frac{1}{z+1}$. This satisfies your hypotheses clearly.
Now the circle is $\gamma(t)=e^{it}, \ \ t \in[0, 2\pi)$, and so the image of this is just the image of $$T(\gamma(t))= \frac{1}{e^{it}+1}$$
Plotting this in Wolfram, the real part is always $1/2$, but the imaginary part is non-linear, so this is not a straight line.
Here's the plot.
While your observation is true, that every Mobius transformation is a composition of these functions, this doesn't mean it must take the circle to a straight line. This particular Mobius transformation is a composition of a translation with an inversion, and so it's image could a priori be a circle or a line, as inversions do not have to have to take circles to lines.