Assuming the Axiom of Choice, an element lies in the Jacobson radical exactly when one minus any multiple is a unit

If $R=0$, then $J(R)=0$ by [L1], the only element is $x=0$, and $1-rx=0$ is the identity element of the zero ring, hence a unit. So the statement holds in this case.

Assume $R\mathrm{\ne }0$ and first suppose $x\in J(R)$. Fix $r\in R$. If $1-rx$ lay in a maximal ideal $\mathfrak{m}$, then [L1] gives $x\in \mathfrak{m}$, hence also $rx\in \mathfrak{m}$, so $1=(1-rx)+rx\in \mathfrak{m}$, impossible. Thus $1-rx$ lies in no maximal ideal.

Conversely, suppose $1-rx$ is a unit for every $r\in R$ and that $x\mathrm{\notin }J(R)$. By [L1], choose a maximal ideal $\mathfrak{m}$ with $x\mathrm{\notin }\mathfrak{m}$. Then the ideal $\mathfrak{m}+(x)$ strictly contains $\mathfrak{m}$, so maximality from [L3] gives $\mathfrak{m}+(x)=R$. Thus $1=a+rx$ for some $a\in \mathfrak{m}$ and $r\in R$, so $1-rx=a\in \mathfrak{m}$. But an element of a proper ideal cannot be a unit, contradicting the hypothesis. Therefore $x\in J(R)$.

Still under the hypothesis of step 1.2, if $1-rx$ were not a unit, then the principal ideal it generates would be proper, so [L2] would place it in a maximal ideal, contradicting step 1.2. Hence $1-rx$ is a unit for every $r$.

Steps 1.1, 2.1, and 1.3 prove the equivalence.