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.)

Second try. Victor points out that my original application of super-rigidity was faulty. This edit is to remove the false claim and leave only the lemma I believe. While I'm at it, I have added some stuff I don't understand.

Let $H$ and $G$ be algebraic groups of units of division algebras over $\mathbb Q$ (because super-rigidity is of limited use over bigger fields). If $f\colon H(\mathbb Q)\to G(\mathbb Q)$ is an isomorphism of abstract groups, then super-rigidity implies that for all $S$, $H(\mathbb Z[S^{-1}])$ and $G(\mathbb Z[S^{-1}])$ contain finite index subgroups identified by $f$. One can't ask for more, since the $S$-integral points of $H$ and $G$ are only well-defined up to finite index. These groups are insensitive to inclusion of the ramified places in $S$. 

I don't know how much this buys us.

Perhaps one can use the congruence subgroup property to reconstruct the local points of the group? I believe that the CSP is conjectured but open for these groups (specifically, that there should be a finite congruence kernel, perhaps for large $S$). Worse, I'm not sure what it says, it particular whether it allows reconstruction of the compact factors. (If this does work, then the compact factors would be the profinite completion of the rational points and one would not need the first paragraph. Maybe this should have been a separate answer.)

If we can reconstruct the compact local factors from the ($S$-)integral points, and thus from the rational points, we can reconstruct the local invariants, up to a sign (the local version of this question, addressed by Victor and Boyarsky). This would not be enough to show that the two discrete groups of rational points are isomorphic as algebraic groups, but it would be enough to show that some examples of groups of rational points are not isomorphic. A division algebra with local invariants $\frac15$ and $\frac45$ has non-isomorphic factors to one with $\frac25$ and $\frac35$. But a division algebra with invariants $(\frac 15,\frac 25,\frac 35,\frac 45)$ is not isomorphic or opposed to one with invariants $(\frac 45,\frac 25,\frac 35,\frac 25)$, even though locally it is always isomorphic or opposed. Thus, such a method could not distinguish their unit groups.

Proof of the claim in the first paragraph that isomorphism of rational points must preserve integral points: Super-rigidity says that maps of lattices extend to maps of their ambient locally compact groups. It requires that the $S$-rank of the source is least $2$, where the $S$-rank is the sum over places in $S$ of the local ranks. Under Kevin's assumptions, we already have rank $2$ from the infinite place; in general we would need to use large $S$ to draw conclusions about small $S$. $H(\mathbb Q)$ is not finitely generated, but it is filtered by groups of $S$-integers, which are finitely generated. Thus an isomorphism $f\colon H(\mathbb Q)\cong G(\mathbb Q)$ 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 restricts to be $f$ on a finite index subgroup of $H(\mathbb Z)$. That is, $f$ preserves the integral points. Similarly for the $S$-integral points.