I believe that showing that every element of $\text{Mob}^+(\mathbb{H})$ is the product of an even number of inversions is quite straightforward (barring a few lemmas here and there). I do not; however, know how I should go about showing that every element of $\text{Mob}^+(\mathbb{H})$ is the product of two inversions?
For your reference:
An element $f$ of $\text{Mob}^-(\mathbb{H})$ having the form: $f(z) = \frac{a \overline{z} + b}{c \overline{z} + d}$ (here where $a, b, c, d \in \mathbb{R}$ and $a d - b c = -1$), is said to be an inversion iff $a + d = 0$.
You should have as a lemma, or can easily prove, that an element of $\mathrm{Mob}^-(\mathbb H)$ with a fixed point in the upper half-plane is an inversion, and that an element of $\mathrm{Mob}^+(\mathbb H)$ with two fixed points in the upper half-plane is the identity.
Given any element $f \in \mathrm{Mob}^+(\mathbb H)$, can you find inversions $g, h \in \mathrm{Mob}^-(\mathbb H)$ such that $g \circ f$ has a fixed point and $h \circ g \circ f$ has two?