Does homeomorphic and isomorphic always imply homeomorphically isomorphic? Let $(G,\cdot,T)$ and $(H,\star,S)$ be topological groups such that

$(G,T)$ is homeomorphic to $(H,S)$ and $(G,\cdot)$ is isomorphic to $(H,\star)$.
Does it follow that $(G,\cdot,T)$ and $(H,\star,S)$ are isomorphic as topological groups?

If no, what if they are both Hausdorff?  What if they are both Hausdorff and two-sided complete?
 A: I see above counter-examples in this thread. On the other hand, in the Abelian case there are positive results, which go way further than the question. In the case of (general) solenoids, they themselves are their own first homology groups with coefficients in $S^1$. It follows that when two of them are homeomorphic (or just homologically equivalent) then they are isomorphic as topological groups--see Karol Borsuk and myself, Fund. Math. 1970. This line was developed much further in later papers by James Keesling.
The following five classifications of (general) solenoids are identical: (i) homological (with coefficients in $S^1$; i.e. cohomological with coefficients in $Z$; for Čech theories), (ii) shape-theoretical, (iii) homotopic, (iv) topological, (v) as topological groups.
A: Sorry for the necromancy, but the following was too cute to resist:
Let 
$A$ be $Z_4$ with the discrete topology
$B$ be $Z_4$ with the indiscrete topology
$C$ be $Z_2 \times Z_2$ with the discrete topology
$D$ be $Z_2 \times Z_2$ with the indiscrete topology.
Then $A \times D$ is not isomorphic to $B \times C$ as a topological group, but the the underlying spaces are homeomorphic and the groups are isomorphic.
A: The Banach spaces $c_{0}$ and $\ell_{2}$ may be viewed as isomorphic
Abelian groups, that are also homeomorphic (due to Kadec). Still,
they are clearly not isomorphic as topological groups.
A: Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.
EDIT: Or the same thing with $\mathbb Z$ instead of $\mathbb Q$.
A: Another example: there are uncountably many different (count-ably based) abelian pro-$p$ groups isomorphic to the product of all the cyclic $p$-groups. Each of these must be homeomorphic to the Cantor set. 
[This is in my transfer report, and shortly to be in a pre-print on the arxiv.]
A: The 2-adic rationals $\mathbb{Q}_2$ and the 3-adic rationals $\mathbb{Q}_3$ are homeomorphic, because each one is a countable disjoint union of Cantor sets.  They are also isomorphic as groups if you assume the axiom of choice, because they are both fields of characteristic 0 and therefore vector spaces over $\mathbb{Q}$ (of the same cardinal dimension).  However, the 2-adic integers $\mathbb{Z}_2$ are a compact subgroup of $\mathbb{Q}_2$ in which every element is infinitely divisible by 3.  On the other hand, in $\mathbb{Q}_3$, any non-trivial sequence $x, x/3, x/9, \ldots$ is unbounded in the complete metric, and is therefore not contained in a compact subgroup.

Keith Conrad asks whether these is an example without the axiom of choice, and Jason De Vito asks whether there is an example using Lie groups.  In fact, there is a cheap example using disconnected Lie groups.  Let $G$ and $H$ be two connected Lie groups that are homeomorphic but not isomorphic.  For instance, abelian $\mathbb{R}^3$, the universal cover $\widetilde{\text{SL}(2,\mathbb{R})}$, and the Heisenberg group of upper unitriangular, real $3 \times 3$ matrices are all homeomorphic, but not isomorphic.  If $G'$ and $H'$ are $G$ and $H$ with the discrete topology, then $G' \times H$ and $G \times H'$ are explicitly isomorphic and explicitly homeomorphic.  But they are not continuously isomorphic, because the connected component of the identity is $G$ for one of them but $H$ for the other one.
