Let $R \neq 0$ be a ring which may not be commutative and may not have an identity. Suppose $R$ satisfies the following conditions.
1) $a^2 = a$ for every element $a$ of $R$.
2) $ab \neq 0$ whenever $a \neq 0$ and $b\neq 0$.
Is $R$ isomorphic to the field $\mathbb{Z}/2\mathbb{Z}$?
On
Let $a \neq 0$ be an element of $R$. It suffices to show that $a$ is an identity. So we would like to show first that $ab = b$ for any element $b$ of $R$. It suffices to show that $a(ab) = ab$ thanks to the condition 2). But this is obvious thanks to the condition 1). Similiarly we can show that $ba = b$.
Assume that $a,b\neq0$. We have $$ 0=abab-ab=(aba-a)b, $$ so $a=aba$. Therefore $$ 0=a-aba=a^2-aba=a(a-ba). $$ So we also see that $a=ba$. Repeating the dose once more $$ 0=ba-a=ba-a^2=(b-a)a\implies b=a. $$ Thus your rng has only one non-zero element. As there are no zero-divisors that non-zero element is an identity, and it follows that $R\cong\mathbb{Z}/2\mathbb{Z}$.