for a ring $R$ with unity , let $U(R)$ denote the group of units of $R$ . Now there are lots of finite commutative rings, of arbitrarily high order, with exactly one unit ; indeed $U(R)=1$ for a finite commutative ring $R$ iff $a^2=a , \forall a \in R$ . Incidentally , I couldn't find any finite non-commutative ring with exactly one unit; matrix rings don't seem to work.
So my question is : Does there exist a finite non-commutative ring with unity having exactly one invertible (unit) element ?
Small remark : Note that such a ring must have characteristic $2$
Best Answer
This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity must be a Boolean ring, hence commutative. The proof given below is based on a result by Melvin Henriksen. It doesn't rely on the Artin-Wedderburn Theorem and turns out to be fully elementary.
Following Melvin Henriksen, we call $R$ a UI-ring if $R$ has an identity element $1$ and $ab = ba = 1$ for $a,b \in R$ implies $a = b = 1$.
We have
The commutative case mentioned in OP's question was solved by P. M. Cohn [2, Theorem 3], should $R$ be finite or infinite:
Addendum. I discovered this preprint of Rodney Coleman (2013) in which OP's question was both asked and answered.
[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.