How far can we go with group isomorphisms?

You need to use the correct notion of isomorphism here: namely, you need to choose an isomorphism between $G$ and $T$ which sends $H_i$ to $S_i$. Such an isomorphism need not exist given only that $H_i$ and $S_i$ are abstractly isomorphic.

In other words, the problem here is that $H_1 H_2$ is not a "property of a group"; it depends on three things, namely a group $G$ and two subgroups $H_1, H_2$ of it, and the correct notion of isomorphism for this situation depends on all three of these pieces of information.

$\newcommand{\ZZ}{\mathbb{Z}}$ The groups may be isomorphic, but when you write something like $H_1 H_2$, you're omitting some information. To talk about this object, you need to know about their parent group $G$, and how $H_1$ and $H_2$ embed into it. And if the way they do so is different from how $S_1$ and $S_2$ embed into $T$, then you'll get a different group.

Here's an (admittedly simple) example. Let $G = T = \ZZ_2 \times \ZZ_2 = \{ e, a, b, ab \}$. Let $H_1 = H_2 = S_1 = \{ e, a \}$ and $S_2 = \{ e, b \}$. Clearly, $H_1 \cong S_1$ and $H_2 \cong S_2$. But $H_1 H_2 = H_1$ and $S_1 S_2 = G$, and these aren't even the same size, much less isomorphic.

This can fail with certain other group properties as well. As a (pretty trivial) example, think about the example above, and $H_1 \cap H_2$ vs $S_1 \cap S_2$. Another property that fails is quotient groups. I think products of groups will always work out though.

One way of thinking of this is that $H_1$ and $H_2$ aren't subgroups of $G$; they're just separate groups that come with an inclusion map, $\iota_i : H_i \to G$. This map is injective, and just sends an element in $H_i$ to its 'copy' in $G$. These maps characterize how $H_i$ is contained in $G$. So if you want to show $H_1 H_2$ and $S_1 S_2$ are isomorphic, you need to find isomorphisms $\alpha: H_1 \to S_1$, $\beta : H_2 \to S_2$, and $\gamma : G \to T$ such that these inclusion maps can "come along for the ride".

As a picture (commutative diagram), you need to find $\alpha$, $\beta$, and $\gamma$ such that it doesn't matter how you follow the arrows. $$ \require{AMScd} \begin{CD} H_1 @>{\iota_1}>> G @<{\iota_2}<< H_2 \\ @V{\alpha}VV @V{\gamma}VV @V{\beta}VV \\ S_1 @>{\iota_3}>> T @<{\iota_4}<< S_2 \end{CD} $$

With the example above, let all three isomorphisms just be the equality map. Then this diagram does not have the desired property. Here's the problematic area: look at $a \in H_2$. If you follow $\iota_2$ to $G$, you get $a$. Following $\gamma$ will give you $a \in T$. But if you follow $\beta$ first, you get $b \in S_2$, and then $\iota_4$ will send you to $b \in T$. We say the diagram "fails to commute".