The question is from an exercise (2.13) in Introduction to Complex Analysis by H.A. Priestley. Before writing this I did check all the questions that might have the answer, but where proofs were given I wasn't able to follow them. (Sorry—self studying beginner here).
So I went back to the inverse points form of a circle in $\mathbb{C}$.
$$|z-\alpha|=\lambda|z-\beta|$$
It is straightforward to show (and Priestley does this) that the condition for $\alpha$ and $\beta$ to be inverse points for
$$|z-a|=r$$ is that $$(\alpha-a)(\overline{\beta-\alpha})=r^2$$
In the case of the unit circle this gives: $\,\alpha\bar{\beta}=1\;$ie one point is the inverse of the conjugate of the other. Which gives the inverse point equation as
$$\left|\frac{z-\alpha}{z-\frac{1}{\bar\alpha}}\right|=\lambda$$ but the circle goes through (1,0) so $$\left|\frac{1-\alpha}{1-\frac{1}{\bar\alpha}}\right|=\lambda=\left|\frac{\bar{\alpha}(1-\alpha)}{\bar{\alpha}-1}\right|$$
and as $|1-\alpha|=|\bar{\alpha}-1|,\;\;$$\lambda=\bar{\alpha}\;$ which gives for the unit disc the equation $$\left|\frac{z-\alpha}{\bar{\alpha}z-1}\right|\lt1$$
As the equation for the unit disc is also $|z|\lt1\;\;$ the transformation $f(z)= $$\left(\frac{z-\alpha}{\bar{\alpha}z-1}\right)$ maps D(0;1) onto D(0;1).
Finally, my question: does having derived this Mobius transformation from the general circle in $\mathbb{C}\;$ constitute a (non-rigorous) proof that all transformations that map D(0;1) onto D(0;1) are of this form or a rotation of it, ie $\;\;e^{i\lambda}\frac{z-\alpha}{\bar{\alpha}z-1}$?