This is close to the famous paper of Borel and Tits, Homomorphismes ``abstraits'' de groupes algébriques simples, Ann. of Math. (2) 97 (1973), 499 – 571 MR0316587 (47 #5134). Unfortunately, one cannot appply their strongest result (abstract homomorphism $\implies$ algebraic homomorphism composed with a field embedding, valid for general infinite fields) because $D$ is anisotropic. Nonetheless, in 9.13 (i) they prove that if $G$ is a semisimple algebraic group over a local field $k\ ({char } k=0)$ and $G'$ is a reductive algebraic group over a local field $k'$ and $G'$ has no nontrivial complex factor, then every abstract homomorphism $G(k) \to G'(k')$ is continuous. Consequently, if $D^*_v \simeq D'^*_v$ as abstract groups then $SL_1(D_v) \simeq SL_1(D'_v)$ as topological groups and I think this implies that $D_v \simeq D'_v$. If this holds for all $v$ then $D \simeq D'.$

This is close to the famous paper of Borel and Tits, Homomorphismes ``abstraits'' de groupes algébriques simples, Ann. of Math. (2) 97 (1973), 499 – 571 MR0316587 (47 #5134). Unfortunately, one cannot appply their strongest result (abstract homomorphism $\implies$ algebraic homomorphism composed with a field embedding, valid for general infinite fields) because $D$ is anisotropic. Nonetheless, in 9.13 (i) they prove that if $G$ is a semisimple algebraic group over a local field $k\ ({char } k=0)$ and $G'$ is a reductive algebraic group over a local field $k'$ and $G'$ has no nontrivial complex factor, then every abstract homomorphism $G(k) \to G'(k')$ is continuous. Consequently, if $D^*_v \simeq D'^*_v$ as abstract groups then $SL_1(D_v) \simeq SL_1(D'_v)$ as topological groups and I thought¹ that this implied that $D_v \simeq D'_v$. If that holds for all $v$ then $D \simeq D'.$²

Footnotes
¹ Wrongly, see Kevin's and BCnrd's examples in the comments.
²True but irrelevant. For Kevin's question, it would be sufficient to find a single pair of $D$ and $D'$ with abstractly non-isomorphic unit groups, and the conclusion is too strong in view of the previous comment.

This is close to the famous paper of Borel and Tits, Homomorphismes ``abstraits'' de groupes algébriques simples, Ann. of Math. (2) 97 (1973), 499 – 571 MR0316587 (47 #5134). Unfortunately, one cannot appply their strongest result (abstract homomorphism $\implies$ algebraic homomorphism, valid for a general infinite field) because $D$ is anisotropic. Nonetheless, in 9.13 (i) they prove that if $G$ is a semisimple algebraic group over a local field $k\ ({char } k=0)$ and $G'$ is a reductive algebraic group over a local field $k'$ and $G'$ has no nontrivial complex factor, then every abstract homomorphism $G(k) \to G'(k')$ is continuous. Consequently, if $D^*_v \simeq D'^*_v$ as abstract groups then $SL_1(D_v) \simeq SL_1(D'_v)$ as topological groups and I think this implies that $D_v \simeq D'_v$. If this holds for all $v$ then $D \simeq D'.$

This is close to the famous paper of Borel and Tits, Homomorphismes ``abstraits'' de groupes algébriques simples, Ann. of Math. (2) 97 (1973), 499 – 571 MR0316587 (47 #5134). Unfortunately, one cannot appply their strongest result (abstract homomorphism $\implies$ algebraic homomorphism composed with a field embedding, valid for general infinite fields) because $D$ is anisotropic. Nonetheless, in 9.13 (i) they prove that if $G$ is a semisimple algebraic group over a local field $k\ ({char } k=0)$ and $G'$ is a reductive algebraic group over a local field $k'$ and $G'$ has no nontrivial complex factor, then every abstract homomorphism $G(k) \to G'(k')$ is continuous. Consequently, if $D^*_v \simeq D'^*_v$ as abstract groups then $SL_1(D_v) \simeq SL_1(D'_v)$ as topological groups and I think this implies that $D_v \simeq D'_v$. If this holds for all $v$ then $D \simeq D'.$