No. Here is a counterexample:
- Take $G = \mathbb{Z}/4\mathbb{Z}$, and $H = \mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$.
- Take any bijection $\phi \colon G \to H$ such that $\phi(0) = (0,0)$.
There are only two proper subgroups of $G$, namely $\{0\}$ and $\langle 2 \rangle$. Both are mapped isomorphically to a subgroup of $H$. But $G \not\cong H$.
So you will need more assumptions. Maybe the requirement that $\phi$ induces a bijection on the lattices of proper subgroups of $G$ and $H$?