Time for a second attempt, here's hoping for the best. Algebras will be strictly non-unital, i.e. they have no global identity element.

**Claim 1:** Let $R$ be a $\mathbb{Z}/2$-algebra with the *EV property*: for every non-zero element $a \in R$ there is a non-zero $v \in R$ such that $av = v$. Then the equation $xy - x + y = 0$ has no solution in $R$ besides $x = y = 0$.

*Proof.* If $v \neq 0$ is such that $yv = v$, then $(xy - x + y)v = xyv - xv + yv = xv - xv + v = v \neq 0$ for any $x$, so $y$ must be $0$ and hence so must $x$. $\square$

**Claim 2:** Let $R$ be a $\mathbb{Z}/2$-algebra with the EV property and let $I$ be the two-sided ideal of $R[x]$ generated by $rx - r$, for all $r \in R$, and $x^2 - x$. If $S = R[x]/I$ then:

- $R$ embeds in $S$, and in particular any non-trivial zero-divisor in $R$ remains a non-trivial zero-divisor in $S$;
- $x \neq 0$ in $S$ and $sx = s = xs$ for all $s \in S$;
- if $0 \neq a \in R$, then $a + x$ is a zero divisor and not a unit in $S$.

In particular, $S$ is a ring with unit $x$.

*Proof.* The first two observations follow from the definition of $I$: every $p \in I$ is either $0$ or has degree at least $1$, so $I \cap R = 0$, and we don't have $x \in S$. Notice first that every element of $S$ is represented by something of the form $a$ or $a + x$ for $a \in R$. If $0 \neq v \in R$ is such that $av = v$, then $(a + x)v = av + xv = v + v = 0$. If an inverse to $a + x$ exists it must be of the form $b + x$ by degree reasons, since $(a + x)b = ab + b \neq x$. But if $(a + x)(b + x) = x$ then we must have $ab + a + b = 0$ in $R$, which forces $a = b = 0$ by Claim 1. $\square$

Now we just need to find a $\mathbb{Z}/2$-algebra that is non-commutative, only has zero divisors, and satisfies the EV property. Let's ignore the zero-divisors for now. Take $A_0$ to be a non-commutative $\mathbb{Z}/2$-algebra, for example the finite rank endomorphism algebra $\mathrm{End}^{\mathrm{f.r.}}((\mathbb{Z}/2)^{\mathbb{N}})$. Let $V_0 = \{\theta_a \mid a \in A_0 \setminus \{0\}\}$ be a new set of variables, $K_0 = \langle a \theta_a - \theta_a \mid a \in A_0 \rangle$ a two-sided ideal of $A_0[V_0]$, and $A_1 = A_0[V_0]/K_0$. Then the subring $A_0 \subset A_1$ has the EV property in $A_1$ since $a \theta_a = \theta_a$ for all $a \in A_0 \setminus \{0\}$ and $\theta_a \neq 0$ since $A_0 \cap K_0 = 0$. Proceed by induction to obtain $A_n \subset A_{n+1}$ with $A_n$ having the EV property within $A_{n+1}$. The union $A = \bigcup_{n = 0}^\infty A_n$ is a non-commutative $\mathbb{Z}/2$-algebra with the EV property. To introduce zero divisors, let $R = \bigoplus_{i = 1}^\infty A$ be the set of sequences of $A$ with only finitely many non-zero entries with the usual pointwise operations. We immediately get zero-divisors for every element (given $r \in R$, take $s$ with disjoint support from $r$; then $rs = 0$), and $R$ still has the EV property: given $r = (a_1, a_2, \dotsc, a_k, 0, 0, \dotsc)$, by the EV property in $A$ can choose $v_i \in A$ such that $a_i v_i = v_i$ (if any of the $a_i$ are zero, chose $v_i = 0$) and so for $s = (v_1, v_2, \dotsc, v_k, 0, 0, \dotsc)$ we get $rs = s$; $s$ is forced to be $0$ only when $r$ is.

And there we go! Claim 2 tells us that the ring $S$ constructed from $R$ has all the desired properties: it is unital, all of its non-unit elements are zero-divisors (the ones of the form $a$ are already zero-divisors since they are in $R$), and it's non-commutative by construction. Moreover, a formal consequence of there not being non-zero nilpotents is that every zero-divisor is a *two-sided* zero-divisor: if $ab = 0$ then $(ba)^2 = baba = 0$ and so $ba = 0$ as well.

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