No. Consider the subring $R\subseteq\prod_{n\geq 1}\mathbb R[x]/(x^n)$ consisting of sequences $(a_n)$ whose constant terms converge to some limit. Let $s=(x,x,x,\dots)$. Then $s$ is nonnilpotent, since $s^n$ has nonzero coefficient in $\mathbb R[x]/(x^{n+1})$.

Now every idempotent $e$ in $R$ is equal to $1$ is equal to $0$ or $1$ in each coordinate, and is eventually constant. Replacing $e$ with $1-e$ if necessary, assume $e=(e_n)$ satisfies $e_n=0$ for $n>N$. Then $s^Ne=0$.

However, for $e=(1,\dots,1,0,0,\dots)$ with $n$ ones we have $s^{n-1}e\neq 0,s^{n-1}(1-e)\neq 0$, so we cannot pick one exponent for all idempotents.

No. Consider the subring $R\subseteq\prod_{n\geq 1}\mathbb R[x]/(x^n)$ consisting of sequences $(a_n)$ whose constant terms converge to some limit. Let $s=(x,x,x,\dots)$. Then $s$ is nonnilpotent, since $s^n$ has nonzero coefficient in $\mathbb R[x]/(x^{n+1})$.

Now every idempotent $e$ in $R$ is equal to $0$ or $1$ in each coordinate, and is eventually constant. Replacing $e$ with $1-e$ if necessary, assume $e=(e_n)$ satisfies $e_n=0$ for $n>N$. Then $s^Ne=0$.

However, for $e=(1,\dots,1,0,0,\dots)$ with $n$ ones we have $s^{n-1}e\neq 0,s^{n-1}(1-e)\neq 0$, so we cannot pick one exponent for all idempotents.