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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.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.

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.

This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity is 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]:

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.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.

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.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.

This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity is 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 $R$ 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]:

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.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.

This answer presents an alternate proof of users' negative answer by proving directly that a finite ring whose only unit is its identity is 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]:

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.

[1] P. M. Cohn, "Rings of zero-divisors", 1984.
[2] M. Henriksen, "Rings with a unique regular element", 1989.