Definition. Call two elements of a commutative ring associates iff each divides the other. Call them strong associates if there exists a unit that can be multiplied by the first to yield the second. (In an integral domain, these are equivalent.)
Every element of a commutative ring is the strong associate of at most one idempotent (exercise). Now consider the commutative ring $\mathbb{Z}/6\mathbb{Z}$. Its idempotents are $\{1,3,4\}.$ Hence not every element of this ring is idempotent. However, every element is the strong associate of some idempotent; $2$ is a strong associate of $4$ (since $2 \times 5 = 4$ and $5$ is a unit) and $5$ is a strong associate of $1$ (since $5$ is a unit.)
However, there exist rings in which some elements are not associates of an idempotent, not even weakly. An example is $\mathbb{Z}/n^2\mathbb{Z}$ whenever $n \geq 2.$ Observe that $n$ is not the associate of an idempotent in this ring, not even weakly. Because if it were, then $n^2$ would be an associate of this idempotent, hence $0$ would be an associate of this idempotent, hence this idempotent would be $0$, hence $n$ would be an associate of $0$, hence $n$ would be $0$.
Question. For which $n \in \mathbb{N}$ is it the case that every element of $\mathbb{Z}/n\mathbb{Z}$ is the strong associate of an idempotent? Precisely the square-free ones, perhaps?
(Side question: is there a name for those commutative rings in which every element is the strong associate of some idempotent? These generalize both fields and Boolean rings simultaneously.)
By using CRT we find that the number of elements in $\mathbb Z/n\mathbb Z$ which are products of an idempotent by an invertible is $N=[1+p_1^{k_1-1}(p_1-1)]\cdots[1+p_r^{k_r-1}(p_r-1)]$, where $n=p_1^{k_1}\cdots p_r^{k_r}$. (Recall that the idempotents in $\mathbb Z/p^k\mathbb Z$ are trivial, and the number of invertibles is $p^k-p^{k-1}$.)
The question reduces to when $N=n$. This clearly happens iff $n$ is square-free.