Isomorphism of two groups on the same set via the knowledge of subgroups 
Let $(H,∘)$ and $(H,∗)$ be two groups and $S⊆H$. Let $\langle S\rangle_{(H,\circ)}$ and $\langle S\rangle_{(H,\ast)}$ be the subgroups generated by $S$ in $(H,∘)$ and $(H,∗)$ respectively. If $\langle S\rangle_{(H,\circ)}=\langle S\rangle_{(H,\ast)}$ for all $S⊆H$, are they isomorphic?

I had some previous discussion regarding this question in the Mathematics chatroom (the details can be seen here) but other than that I couldn't progress at all regarding this question. 
The only observation (admittedly trivial) that I have been able to make is that the identity elements of the groups are identical. This follows by taking $S=\emptyset$.
 A: Just for the record, here is the example that I mentioned in my comment. I am still not planning to write down a proof: I labouriosly checked that it works by computer.
As in my answer to this question, we take
$$G = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^5 \rangle,$$
$$H = \langle x,y,z \mid x^{11}=y^{11}=z^5=1, xy=yx, x^z=x^4, y^z=y^3 \rangle.$$
These are $\mathtt{SmallGroup}(605,5)$ and $\mathtt{SmallGroup}(606,6)$ in the small groups database.
Observe that we can write the elements of both $G$ and $H$ as
$$\{ x^iy^j : 0 \le i \le 10,\,0 \le j \le 10\} \cup
\{ x^i(y^j z)^k: 0 \le i \le 10,\,0 \le j \le 10,\, 1 \le k \le 4 \}, $$
and we use this representation to identify the elements of the underlying sets of $G$ and $H$. We do it this way to make the subgroups of order $5$ correspond in the two groups.
The subgroups of order $11$  are all contained in the normal subgroup $\langle x,y \rangle$ and obviously correspond. So it really only remains to check that the $22$ subgroups of order $55$ correspond. Note that half of these contain $\langle x \rangle$ and the other half contain $\langle y \rangle$. I convinced myself that they did correspond and then checked it by computer.
Added: I think that insisting that the two groups should have the same underlying sets is confusing. Here is an equivalent statement of the problem, which I find easier to work with.
Do there exist non-isomorphic group $G$ and $H$ such that there is a bijection $\phi:G \to H$ that induces a bijection between the subgroups of $G$ and the subgroups of $H$. That was the version that I verified by computer.
