Before the issue will make some definitions. If $\operatorname{End}(A)$ is the ring of endomorphisms of group $A$. For any $a,b \in A$, define a two-sided multiplication $_aM_b \in \operatorname{End}(A)$ by $_aM_b(x) = axb$. With $M(A)$ we denote the set of all elements in $\operatorname{End}(A)$ that can be written as a finite sum of two-sided multiplications $_aM_b$. An element $f \in M(A)$ is thus of the form $f:x \mapsto \sum_i a_ixb_i$; note that $M(A)$ is a subring of $\operatorname{End}(A)$.
If $A$ is a simple unital ring then the associated division ring $\operatorname{End}({}_{M(A)}A)$ is isomorphic to $Z(A)$, the center of $A$.
I'll write endomorphisms of $_{M(A)}A$ on the right.
Let $f\colon A\to A$ be an endomorphism of $_{M(A)}A$. If $a,b\in A$, we have $({}_aM_bx)f={}_aM_b(x)f$, for every $x\in A$, that means $$ (axb)f=a(x)fb $$ In particular, for $b=1$, this implies $$ (ax)f=a(x)f $$ and so $f$ is an endomorphism of $_AA$ and so we can identify it with a subring of $A$, via the evaluation at $1$. So, let $r=f(1)$. Then, taking $x=1$ and $a=1$, $$ (1b)f=(1)fb $$ or $$ (1b)f=rb $$ But $1b=b1={}_bM_11$ and so $$ (1b)f=(b1)f=({}_bM_11)f={}_bM_1(1)f=br $$ This proves that $r\in Z(A)$.
That, for $r\in Z(A)$, the map $x\mapsto xr$ defines an endomorphism of $_{M(A)}A$ is easy to show.
Note that simplicity of $A$ is not needed. It is for stating that $Z(A)$ is a division ring, hence a field, because $_{M(A)}A$ is a simple module when $A$ is a simple ring (the submodules are two-sided ideals).