Do these properties of a countable abelian group guarantee a Prüfer subgroup?

Yes, it must. And $G$ doesn't need to be countable.

Let $H$ be the $p$-primary component of the torsion subgroup of $G$. Then the natural map $H/pH\to G/pG$ is injective, so $H$ also satisfies (1), and clearly satisfies (2). So, replacing $G$ by the subgroup $H$, we shall assume that $G$ is a $p$-group.

Let $X$ be a finite subset of $G$ that generates $G/pG$, so $$G=\langle X\rangle + pG.$$

For some $n$, $p^nx=0$ for all $x\in X$, and so $$p^nG=p^n\langle X\rangle + p^{n+1}G=p^{n+1}G.$$

So $p^nG$ is divisible, and by (2) is nonzero. And a divisible abelian $p$-group is a direct sum of copies of the Prüfer $p$-group.