In a problem I've just completed, I show that we have the isomorphism $SL(2,\mathbb{C})/\{-I,I\} \cong \mathcal{M}$ where $SL(2,\mathbb{C})$ is the group of 2x2 complex unimodular matrices, $\{-I,I \}$ are the identity and minus the identity matrices in that group, and $\mathcal{M}$ is the set of Mobius transformations of the complex plane. The author goes on to say that $SL(2,\mathbb{C})/\mathbb{Z}_2 \cong \mathcal{M}$.
My question is, does it even make sense to talk about $SL(2,\mathbb{C})/\mathbb{Z}_2$ when $\mathbb{Z}_2$ is not strictly speaking a (normal) subgroup (or even subset) of $SL(2,\mathbb{C})$? Or is this just a manner of speaking since clearly $\mathbb{Z}_2 \cong \{-I,I\}$?
Yes to your last question. This is a common (slight) abuse. For example we may say $\mathbb{Z}_4 / \mathbb{Z}_2 \cong \mathbb{Z}_2$, even though $\mathbb{Z}_2$ (the set of congruence classes mod $2$) is not literally a subset of $\mathbb{Z}_4$ (the set of congurence classes mod $4$).
Occasionally (but not too often in practice) this can lead to ambiguity. For example, what is $(\mathbb{Z}_4 \times \mathbb{Z}_2) / \mathbb{Z}_2$? Well, it depends on what isomorphic copy of $\mathbb{Z}_2$ in $\mathbb{Z}_4 \times \mathbb{Z}_2$ we're talking about. The answer could be $\mathbb{Z}_4$ or it could be $\mathbb{Z}_2 \times \mathbb{Z}_2$. Context is key in these situations.
For $\mathrm{SL}(2,\mathbb{C})$ there is no ambiguity because there is only one normal subgroup of order $2$ (indeed there is only one element of order $2$).