$G/ \sim$ is the set of $\sim$-equivalence classes in $G$ and $|G/ \sim|$ is the cardinality of $G/ \sim$. $|A| \leq |B|$ means that there is an injective function from $A$ to $B$. $|A| < |B|$ means that $|A| \leq |B|$ and $|A| \neq |B|$.
If $\pi:G \to G/ \sim$ is defined by $\pi (x) = [x]_{\sim}$, then $\pi$ has an injective right inverse $h:G/ \sim \to G$ (by the Axiom of Choice) and $|G/ \sim| \leq |G|$ and $|G| \nless |G/ \sim|$
Is the full axiom of choice needed to prove that $|G| \nless |G/ \sim|$ (just that one part, not necessarily both of $|G/ \sim| \leq |G|$ and $|G| \nless |G/ \sim|$)? If not, what is the weakest form of choice that is needed to prove it?
If the restriction that "$G$ is a group, $N$ is a subgroup of $G$ and $\sim$ is defined by $x \sim y$ iff $x y^{-1} \in N$" was added, does that change the answer?
This is known as the Weak Partition Principle. We don't know all too much about it, but it implies there are no-measurable sets of reals and there are no infinite Dedekind-finite sets.