The Mobius transformation $x=\frac{\mu}{1-\mu}$ converts a semiline ($x\geq 0$) to an interval $0\leq\mu\lt 1$.
The Mobius transformation $x=\frac{\mu}{1+\mu}$ converts a semiline ($0\leq x$) to an interval $-1\lt\mu\leq 0$.
The transformation $x=\frac{\mu}{1-|\mu|}$ converts a line ($x\in\mathrm{R}$) to an interval $-1\lt\mu<1$. But $x(\mu)$ is only $C^{1}$.
Question: Is there a Mobius transformation that converts a line ($x\in\mathrm{R}$) to an interval $-1\lt\mu<1$?
By definition, a Möbius transformation must have the form $$ x \mapsto\frac{ax+b}{cx+d} $$ If $c=0$, the transformation clearly doesn't do what you want.
And if $c=0$ then the function blows up* near $x=-d/c$, so doesn't do what you want either.
*: unless $-d/c=-b/a$, but then it is not a Möbius transformation at all since $ad-bc=0$; it is then constant when $x\ne-d/c$.
In general, no rational function can have different finite limits for $x\to+\infty$ and $x\to-\infty$, which would be necessary to map $\mathbb R$ homeomorphically to $(-1,1)$.