Let $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and let $\mathbb{F}_{1}^{\ast}$ and $\mathbb{F}_{2}^{\ast}$ denote the corresponding groups of units. If $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ are not elementarily equivalent in the language of rings, does it follow that $\mathbb{F}_{1}^{\ast}$ and $\mathbb{F}_{2}^{\ast}$ are not elementarily equivalent in the language of groups? For example, $\mathbb{R} \not\equiv \mathbb{C}$, and $\mathbb{R}^{\ast} \not\equiv \mathbb{C}^{\ast}$ since (for example) the sentence $$ \exists x \ x^4 = 1, x^2 \neq 1 $$ holds in $\mathbb{C}^{\ast}$ and does not hold in $\mathbb{R}^{\ast}$. Similarly, $\mathbb{Q} \not\equiv \mathbb{R}$, and $\mathbb{Q}^{\ast} \not\equiv \mathbb{R}^{\ast}$ since (for example) the sentence $$ \exists x \ x^2 = 1, x \neq 1, \forall y \ \exists z \ (y = z^2) \vee (xy = z^2)$$ holds in $\mathbb{R}^{\ast}$, but not in $\mathbb{Q}^{\ast}$.
It is natural to consider the contrapositive of this problem: is it true that $\mathbb{F}_{1}^{\ast} \equiv \mathbb{F}_{2}^{\ast} \Longrightarrow \mathbb{F}_{1} \equiv \mathbb{F}_{2}$? I suspect that this is not in general true, mainly because there are many well-known examples of groups which are elementarily equivalent but not isomorphic. So it seems plausible that one may use groups $G_{1}$ and $G_{2}$ which are known to satisfy $G_{1} \equiv G_{2}$ and $G_{1} \not\cong G_{2}$ to construct elementarily inequivalent fields $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ such that the underlying multiplicative group of $\mathbb{F}_{1}$ (resp. $\mathbb{F}_{2}$) is $G_{1}$ (resp. $G_{2}$).
For example, it is known that given an abelian group $A$, $A$ has unbounded exponent if and only if $A \equiv A \oplus \mathbb{Q}$ (Lemma A.2.4. in Hodges' "Model Theory"). Therefore, $\mathbb{Q} \oplus \mathbb{Q} \equiv \mathbb{Q}$. But is the additive group $\mathbb{Q}$ the underlying multiplicative group of a field? Is $\mathbb{Q} \oplus \mathbb{Q}$ the group of units of a field?
As early as 1960, Laszlo Fuchs asked which abelian groups can be realized as the underlying multiplicative group of a field. This question remains largely unanswered. However, it is known that:
Theorem: A nontrivial torsion-free divisible abelian group $G$ has infinite rank if and only if $G$ can be realized as the underlying multiplicative group a field (see "Divisible Multiplicative Groups of Fields" by Greg Oman).
Therefore, there exist fields $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ such that $\mathbb{F}_{1}^{\ast} = \mathbb{Q}$ and $\mathbb{F}_{2}^{\ast} = \mathbb{Q} \oplus \mathbb{Q}$. But it is not obvious as to whether or not these fields are elementarily equivalent.
Take $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$. Their unit groups are isomorphic, not just elementary equivalent, by unique prime factorization, but $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ are not elementary equivalent because there exists a solution to $x^2 = 2$ in the latter but not the former.