Let $R$ be a ring with identity (commutative or not), and assume $J(R)$ to be its Jacobson radical. Let $e\in R $ be an idempotent and $x\in J(R)$ be a given fixed element. I am searching for unknown elements $a,b\in J(R)$ so that the equality $eae+(1-e)b(1-e)=x$ does hold.
The left hand side expands as $eae+b-be-eb+ebe$ and I chose alternatives for $a,b$ but no sum of which totals $x$. By finding $a,b$, if any, one could infer that $eJ(R)e+(1-e)J(R)(1-e)=J(R)$.
Thanks for any suggestion!
Consider the ring $R = \left( \begin{smallmatrix} k & k \\ 0 & k \end{smallmatrix} \right)$ of upper triangular $2 \times 2$-matrices over some field $k$, then $J(R) = \left( \begin{smallmatrix} 0 & k \\ 0 & 0 \end{smallmatrix} \right)$ and for $e = \left( \begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right)$ we get $e J(R) e = \{0\}$ and $(1 - e) J(R) (1 - e) = \{0\}$.
This shows that what you are trying to prove does not work in general.
In fact, note that you always have a direct sum decomposition $J(R) = eJ(R) e \oplus e J(R) (1-e) \oplus (1-e)J(R)e \oplus (1-e) J(R) (1-e)$. In the above example, we have $J(R) = e J(R) (1-e)$.
However, if $e \in R$ is a central idempotent, then the above direct sum decomposition reduces to $J(R) = e J(R) e \oplus (1-e) J(R) (1-e)$, since $e (1- e) = 0$.