Finite non-commutative ring with few invertible (unit) elements
I think I have it ; there can be no such non-commutative ring.
Let $x \in J(R)$ , then $1-x$ is a unit of $R$ , so $x=0$ i.e. $J(R)=0$ . Thus $R$ is an artinian semisimple ring , so by Artin-Wedderburn , $R \cong \prod_{i=1}^m M(n_i , D_i) $ , where $D_i$'s are division rings . But $R$ is finite , hence so are $D_i$'s , hence by Wedderburn little theorem, $D_i$'s are fields , so $R \cong \prod_{i=1}^m M(n_i , k_i) $ , where $k_i$'s are fields . Now since $R$ is not-commutative , at least one $n_i$ is more than $1$ , say w.l.o.g. $n_1 \ge 2$ , but then $M(n_1 , k_1)$ has at least $q^{n_1}-1 \ge q^2-1 >1$ many units (where $q=|k_1|$) , so $R$ has more than one unit .
In fact , since $M(n_1,k_1)$ has $\prod_{j=0}^{n_1-1}(q^{n_1} - q^j)$ many units , where $q=|k_1|$ and for $n_1 \ge 2$ , $\prod_{j=0}^{n_1-1}(q^{n_1} - q^j)\ge (2^2-1)(2^2-2)=6$ ; so we get that :
Any finite non-commutative ring with unity and with zero Jacobson radical has at least $6$ units .
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
Claim. A finite ring $R$ with identity is a UI-ring if and only if $R$ consists only of idempotent elements, i.e., $R$ is a Boolean ring. In particular, a finite UI-ring is commutative.
Proof. Assume that $R$ is a UI-ring. Then $R$ is reduced and $2x = 0$ for every $x \in R$. As $R$ is a finite dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$, every element of $R$ is algebraic over $\mathbb{Z}/2\mathbb{Z}$ by the Cayley-Hamilton theorem. Thus $R$ is a Boolean ring by [2, Corollary 2.10], which shows that $R$ is commutative. Assume now that $R$ is a Boolean ring. As any element $x \neq 1$ satisfies $x(1 - x) = 0$, the identity $1$ is the only unit of $R$.
The commutative case mentioned in OP's question was solved by P. M. Cohn [2, Theorem 3], should $R$ be finite or infinite:
Cohn's Theorem Let $R$ be an algebra over a field $F$ without nontrivial units, i.e., the units of $R$ are those of $F$. Then $R$ is a subdirect product of extension fields of $F$, and every element $x$ of $R$ which is not in $F$ is transcendental over $F$, unless $F = GF(2)$ and $x$ is idempotent. If, moreover, $R$ has finite dimension over $F$, then either $R = F$ or $R$ is a Boolean algebra.
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.