 Solution

Suppose $x\in J$ and put $u=1+yx$. If the left ideal $Au$ were proper, it would lie in a maximal left ideal $L$. But $yx\in J\subseteq L$, so $u\in L$ would imply $1\in L$. Thus $Au=A$, and $u$ has a left inverse $v$ with $vu=1$.

A left inverse alone is insufficient in a general operator algebra. Here $v=1-vyx$ differs from the identity by an element of $J$. The same argument supplies a left inverse $w$ of $v$. Then $w=w(vu)=(wv)u=u$, hence $uv=wv=1$. So $u$ is genuinely invertible.

Conversely, if $x\notin J$, part (i) gives a maximal left ideal $L$ with $x\notin L$. Maximality implies $L+Ax=A$, so $1=l+ax$ for some $l\in L,a\in A$. Therefore $1-ax=l\in L$ cannot be invertible, since a proper left ideal contains no invertible element. Taking $y=-a$ disproves the required universal invertibility. We have proved the unit criterion for the Jacobson radical in the requested form: