Assuming the Axiom of Choice, a nonzero commutative ring is local exactly when its nonunits form an ideal, exactly when one of $x$ and $1-x$ is a unit for every $x$

Assume $R$ is local with maximal ideal $\mathfrak{m}$. No element of $\mathfrak{m}$ is a unit. Conversely, if $x$ is a nonunit, then $(x)$ is proper and [F2] places it in a maximal ideal, necessarily $\mathfrak{m}$. Thus the nonunits are exactly $\mathfrak{m}$, proving condition 2.

Assume the nonunits form an ideal $N$. If both $x$ and $1-x$ were nonunits, then [F4] would give $1=x+(1-x)\in N$, contrary to [F3]. Hence condition 3 holds.

Assume condition 3 and let $N$ be the nonunits. By [F3], $0\in N$. If $x\in N$ and $r\in R$, then $rx$ cannot be a unit, since an inverse for $rx$ would make $x$ a unit; also $-x$ cannot be a unit.

If $x,y\in N$ and $u=x+y$ were a unit, then $x{u}^{-1}$ and $y{u}^{-1}=1-x{u}^{-1}$ would both be nonunits, because a unit among either would make $x$ or $y$ a unit. This contradicts condition 3. Thus $x+y\in N$, and [F4] with step 1.3 shows that $N$ is an ideal.

The ideal $N$ is proper because $1\mathrm{\notin }N$. Every proper ideal consists entirely of nonunits, so it is contained in $N$. Hence $N$ is maximal and is the only maximal ideal; by [F1], $R$ is local.