Suppose that every nonempty definable (without parameters) subset of the group $G$ contains an element of finite order.
Must $G$ be elementary equivalent to a countable torsion group?
My first idea is to take the set of definable elements of finite order but the product of two finite order elements need not be of finite order.
This is a nice application of the omitting types theorem.
Let $\varphi_n(x)$ be the formula $x^n = e$ (where of course $x^n$ is an abbreviation for the $n$-fold product of $x$). Let $\Sigma(x) = \{\lnot \varphi_n(x)\mid n\in \omega\}$. An element satisfies the partial type $\Sigma(x)$ if and only if it has infinite order. So we seek a countable model of $\text{Th}(G)$ in which no element satisfies $\Sigma(x)$.
By the omitting types theorem, such a model exists as long as there is no formula $\psi(x)$ such that $\text{Th}(G)\models \exists x\, \psi(x)$ and $\text{Th}(G)\models \forall x\, \psi(x)\rightarrow \lnot\varphi_n(x)$ for all $n\in \omega$. This is exactly the hypothesis you're given: no nonempty definable subset of $G$ contains only elements of infinite order.