This arises from a (very early) exercise in H. A. Priestley's Introduction to complex analysis. Given the transformation $\frac{z-1}{z+1}$ we were to find the invariant (=fixed?) points, which are $\pm$ i, and apply the transformation to various objects. When applied to the real axis, the transformation appeared to return the whole of the complex plane—every $z$ was included. I used the inverse point form for the real axis, ie $|z-i|=|z+i|$. substituted for $z$, and after multiplying through by $z+1$ and dividing through by $1-i$, (not sure whether this is legitimate...) obtained$$|z-1|=|iz-i|$$ which is true, I think, for all $z$.
Later: I tried the same transformation on $|z|=2$ which doesn't go through $i$ or -$i$ either but it returned a perfectly normal looking circle...which didn't involve the 'fixed points' either.
The Möbius transformation $z \rightarrow f(z)=\frac {z-1}{z+1}$ can be re-written as:-
$z \rightarrow f(z) = \frac{(z-1)(\bar{z} + 1)}{(z+1)(\bar{z} + 1)} = \frac{|z|^2 -1 + 2i \text{Im}(z)}{|z+1|^2}$
If $z$ is on the real axis then $\text{Im}(z)=0$ so $f(z)$ is also real i.e. $f$ maps the real axis to itself.
If $z$ is on the imaginary axis then $|z+1|=|z-1|$ so $|f(z)|=1$ i.e. $f$ maps the imaginary axis to the unit circle.
If $|z|=1$ then $f(z) = \frac{2\text{Im}(z)}{|z+1|^2}$ which is purely imaginary i.e. $f$ maps the unit circle back to the imaginary axis.
$\pm i$ are mapped to themselves and $\infty \rightarrow 1 \rightarrow 0 \rightarrow -1 \rightarrow \infty$. In fact, you can show that $f^4(z)=z$ for all $z$.
In general, a Möbius transformation is a 1-1 mapping of the extended complex plane $C \cup {\infty}$ to itself. So the image of a subset of the complex plane (such as the real axis) cannot be the whole complex plane, otherwise the mapping would not be 1-1.