Take $G = \mathbb{Q}/\mathbb{Z}$; then $\text{End}(G)$ is the profinite integers

$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$

where $\mathbb{Z}_p$ is the $p$-adic integers. The element $\prod_p p$ is neither a unit nor a zero divisor in this ring, and it cannot be written as a product of irreducible elements. This is because the only irreducible elements $x = \prod_p x_p$ are those of the form $x_q = q$ for a fixed prime $q$ and $x_p = 1$ otherwise, and unit multiples of these (there are lots of units), and so the elements that can be written as a product of irreducibles are precisely those where $x_p$ is a unit for all but finitely many $p$.

Take $G = \mathbb{Q}/\mathbb{Z}$; then $\text{End}(G)$ is the profinite integers

$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$

where $\mathbb{Z}_p$ is the $p$-adic integers. The element $\prod_p p$ is neither a unit nor a zero divisor in this ring, and it cannot be written as a product of irreducible elements. This is because the only irreducible elements $x = \prod_p x_p$ are those of the form $x_q = q$ for a fixed prime $q$ and $x_p = 1$ otherwise, and unit multiples of these (there are lots of units), and so the elements that can be written as a product of irreducibles are precisely those where $x_p$ is a unit for all but finitely many $p$, and nonzero otherwise.