Cancellation law in a ring without unity

However, if our ring doesn't contain unity, but the cancellation law holds, then what can we make of the equation $a^2=a$?

In a ring which satisfies cancellation on both sides, $a$ can only be one of two things: $0$, or else the identity for the ring.

Obviously $0$ satisfies $0^2=0$, and if $a$ is nonzero and satisfies $a^2=a$, then $a(ar-r)=(ra-r)a=0$ for arbitrary $r$, whereupon you must conclude that $ar=r=ra$ for all $r$, and that $a$ is the identity.

Of course, depending on your ring, $0$ may be the only element that satisfies $a^2=a$ (like the rng $(X)$ inside the ring $F[X]$.)

What the "paradox" suggests is that in a ring with cancellation and no identity element the only solution to the equation $x^2 = x$ is $x=0$.

If there is a nonzero solution $a$ to that equation then for any $b$ in the ring $$ 0 = (a^2 - a)b = a^2 b - ab $$ so $$ a^2 b = ab . $$ Then cancelling $a$ implies $$ ab = b . $$ The same argument works for right multiplication so $a$ is a (hence the) multiplicative identity.

The usual definition of an integral domain requires an identity element. https://en.wikipedia.org/wiki/Integral_domain.

It's possible for a ring without unity to have no nonzero zero-divisors.

For example, let $R$ be the ideal $(2)$ of $\mathbb{Z}$. Then $R$ is a ring with no multiplicative identity, and no nonzero zero-divisors.

For a finite example, let $R$ be the ideal $(2)$ of $\mathbb{Z_6}$. Then once again, $R$ is a ring with no multiplicative identity, and no nonzero zero-divisors.

For the above examples, given the lack of nonzero zero-divisors, we get a cancellation law:

$\;\;{\small{\bullet}}\;\,$If $ab=ac$ and $a \ne 0$, then $b=c$.

However, as the answers by rschwieb and Ethan Bolker make clear, if a ring $R$ has a nonzero idempotent element $a$ (i.e., an element $a\;$such that $a^2=a$, and $a\ne 0$), then the element $a$ must be a multiplicative identity for $R$.