Let $T$ be a first-order theory of cyclic groups. Even if an abelian group $(G,+)$ satisfy $(G,+)\models T$ there is no reason that $(G,+)$ is a cyclic. (For example, by Löwenheim–Skolem theorem there is uncountable abelian group $G$ that satisfy $T$.)
I tried to find a first-order formula that is true for all cyclic groups, but is false for some abelian group. But I don't know how to find it. Thanks for any help.
To make my comment more formal, the following statement (using multiplicative notation for groups) is true in all cyclic but not in all abelian groups. It is false, for example, in the free abelian group of rank 2, or in the Klein 4-group.
$\forall x,y \in G, \exists z \in G$ such that $z^2=x \vee z^2=y \vee z^2=xy.$
A statement that is true in all 2-generator but not in all 3-generator abelian groups is
$\forall x,y,z \in G, \exists w \in G$ such that $w^2=x \vee w^2=y \vee w^2=z \vee w^2=xy \vee w^2=xz \vee w^2=yz \vee w^2 = xyz.$