Non-commutative ring Local Ring

If you proved that $L$ is in fact a two sided ideal, then congratulations, the difficult (or the more ellusive - at least to me) part is over:

Clearly $U(R) \cap L= \emptyset,$ otherwise $L=R$. That is, $U(R) \subseteq R \setminus L$.

Now suppose that $a \in R \setminus L$. Then $Ra$ is a left ideal not contained in the unique left maximal ideal $L$. That is, it is improper, i.e. $Ra=R$. This means that $ba=1$ for some $b \in R$. Now, we again have $b \notin L$, for if $b \in L,$ then $1=ba \in L$ (because $L$ is a right ideal as well) and $L=R,$ contradiction. So we may repeat the argument for $b$ to get a $c \in R$ with $cb=1$. But now we have

$$a=1\cdot a=(cb)a=c(ba)=c \cdot 1=c,$$ so we have $ab=1$ as well, which proves that $a \in U(R)$.

It's true.

You can show that the set of nonunits of a ring is closed under addition iff the ring is local. (I think it is already here on math.SE but I couldn't find it.)

You can get at it lots of other ways too.

Here's a direct proof. I guess you agree $U(R)\subseteq R\setminus L$.

If $x$ isn't in $L$, then $Rx=R$. But then $yx=1$ implies $xy$ is a nonzero idempotent of $R$. The only nonzero idempotent of a local ring is $1$, so $x$ is a unit.