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. May I suggest the requirement that

>**$(\star)\quad$** $\phi$ induces a bijection between the posets of proper subgroups of $G$ and $H$.

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As remarked in the comments, if $G$ does not have a generating set with $\le 2$ elements, then $\phi$ must be a homomorphism, hence an isomorphism.

If you add the assumption $(\star)$, then cyclic groups are fine as well. The lattice of subgroups is distributive if and only if the group is locally cyclic. If the group is finitely generated, this means that it must be cyclic (see http://planetmath.org/latticeofsubgroups).

*Remark:* I am not yet sure what to do with groups generated by $2$ elements, but I have the feeling those shouldn't be hard now. I do not have anything close to a proof though. Please let me know if you think $(\star)$ makes sense in your setting.