The claim is not true. Suppose that $n= 3$ and take $A=\lbrace E_{1,2}+E_{2,3}, E_{3,2}-E_{2,1}\rbrace$, and let $S$ be the span of $A$.
Note that all matrices from $S$ are nilpotent. However, if we compute $E(X)$, $E^2(X)$ and $E^3(X)$, we see that
the entries from $E(X)$ and $E^3(X)$ are equal up to a scalar $1$ or $2$. In fact, the coefficients $x_{21},x_{12},
x_{23},x_{32}$ of $X=(x_{ij})$ are fixed, i.e., the same for $E(X)$ and $E^3(X)$.
For example, $E(X)=E^3(X)$ for
$$
X=\begin{pmatrix} 0 & 1 & 0 \cr
1 & 0 & 1\cr
0 & 1 & 0
\end{pmatrix}.
$$
So there is no $d$ with $E^d(X)=0$ for all $X$, if I am not mistaken.
The point is that not all subspaces $S$ of $M_{n,n}$ consisiting of nilpotent elements are conjugated to some
linear space of stricly trianguar matrices.

Edit: $A=E_{1,2}$ and $B=E_{2,1}$ is not a counterexample, as pointed out by Abhinav Kumar.