Injective map of abelian group and product of cyclic quotients

No.

Let $A=\mathbb{Q}$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $\prod (A/I)$ is trivial.

To complement @user10354138's answer:

It's true iff $A$ is residually finite.

If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $\mathbf{Q}$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_{\mathrm{div}}\oplus B$, where $A_{\mathrm{div}}$ is its subgroup of divisible elements (this is a vector space over $\mathbf{Q}$), and a residually finite abelian group $B$.