Here's what I can get from super-rigidity for groups over $\mathbb Q$ (over other fields, it buys rather less). It shows that all the local factors are isomorphic and that the isomorphism approximately preserves the $S$-integral points. The isomorphism of local factors gives examples of groups that are not isomorphic. I think integral points should know everything so that isomorphism requires that the groups are algebraically isomorphic, but I'm not sure.

Once we know the local factors are isomorphic, we're lead to the same question for local fields. The original question doesn't really motivate global fields, as opposed to general fields. Local fields should be easier! Boyarsky's comments on Victor's answer addresses the local case: continuous isomorphism must be algebraic; so the algebras are isomorphic or opposed; we can extract the invariant, up to a sign. So if we have two units groups over $\mathbb Q$ that are isomorphic, they have isomorphic local factors, so their local invariants are equal, up to *local* signs. Thus, all $\frac 15$ is not isomorphic to all $\frac 25$, though this doesn't distinguish $(\frac 15,\frac 25,\frac 35,\frac 45)$ from $(\frac 45,\frac 25,\frac 35,\frac 25)$; or distinguish $10$ at $\frac15$ from $5$ at $\frac15$ and $5$ at $\frac45$; or do anything for $p=3$.

**Proof**: Super-rigidity says that maps of lattices extend to maps of their ambient locally compact groups. There's probably a version groups of rational points, which are lattices in adelic groups, but I've never heard. That would immediately give the local factors. But I do know that there's a version for $S$-integral groups. $H(\mathbb Q)$ is not finitely generated, but it is filtered by groups of $S$-integers, which are finitely generated. Thus an isomorphism $H(\mathbb Q)\cong G(\mathbb Q)$ yields a homomorphism $H(\mathbb Z)\to G(\mathbb Z[S^{-1}])$, for some $S$. By super-rigidity, $H(\mathbb Z)$ contains a finite index subgroup on which this extends to a (continuous) homomorphism $H(\mathbb R)\to G(\mathbb R\times \mathbb Q_p\times\ldots)$, which has to land in $G(\mathbb R)$. And similarly back, so we get an isomorphism $H(\mathbb R)\cong G(\mathbb R)$ which is the given isomorphism on a finite index subgroup of the integral points. Similarly, for each $S$, the $S$-integral points are, up to finite index, preserved by the given isomorphism and the local factors are isomorphic. (That proves some kind of adelic/$G(\mathbb Q)$ version of the theorem, but I'm not sure what the statement is. A weak statement is that an isomorphism of $\mathbb Q$-points implies an isomorphism of adelic groups.)