Can one recover $G$ from $G/H$?

Note that the answer that follows resulted from me misinterpreting the question. I understood "recover $G$" to mean recover $G$ up to an isomorphism that preserves the given group structure on $G/H$. But the intention was to recover the group operation on $G$ exactly. Since, whenever $|H|>1$ there is no way of identifying the identity element of $G$, it is only possible to recover $G$ exactly when $|H|=1$.

For any group structure on $H$, is possible that $G = G/H \times H$. So we can only recover $G$ from $G/H$ if there is a unique group structure on $H$, and if the direct product is the only extension.

There is a unique group structure on $H$ if and only if $H$ is finite and has order equal to either $1$ or a prime. If $|H|=1$ then $G$ is determined by $G/H$, so assume that $|G/H|=p$ is prime.

For the extension to be unique, we require first that there is no nontrivial action of $G/H$ on $H$ or equivalently that the commutator quotient $H_1(G/H)$ of $G/H$ is finite and has order coprime to $p-1$. If that holds, then $H \le Z(G)$.

Finally, we require that there is no nonsplit central extension of $H$ by $G/H$, which is equivalent to $H_1(G/H)$ having order coprime to $p$, and the Schur Multiplier $H_2(G/H)$ of $G/H$ being finite and having order coprime to $p$.

Summing up, the conditions for $G$ to be uniquely determined by $G/H$ are that either (i) $|H|=1$; or (ii) $|H|=p$ is finite and prime, $H_1(G/H)$ is finite of order coprime to $p(p-1)$, and $H_2(G/H)$ is finite of order coprime to $p$.

Let $G$ be a group of order two with underlying set $\{a,b\}$ (but I won't tell you which of these is the neutral element). Then $G/G$ is a group with underlying set $\{\{a,b\}\}$ and trivial operation, $\{a,b\}*\{a,b\}=\{a,b\}$. Clearly, we can recover the underlying set $\{a,b\}$ from this. But all we know about $G/G$ is symmetric with respect to swapping $a\leftrightarrow b$. Consequently, with this knowledge alone we cannot determine which of $a,b$ is the neutral element of $G$.