One possibility, of course, is $R=\{0\}$. Assume $1_R\neq 0$.
$R$ has no zero divisors: if $xy=0$ and $x\neq 0$, then $y = 1_Ry = xxy = x(0)=0$.
$R$ is commutative: if $x$ and $y$ are nonzero, then so is $xy$ by the above; hence $(xy)^2 = x^2y^2$; canceling from $xyxy=xxyy$ we get $yx=xy$.
(Of course, the ring satisfies $x^3=x$ for all $x$, so by a famous theorem of Jacobson, the ring is necessarily commutative; but we don't need to call in that heavy cannon to the fray).
Since every nonzero element is invertible, $R$ is a field. Since $x^2-1_R$ has $|R-\{0\}|$ solutions, we have $|R-\{0\}|\leq 2$, so $|R|\leq 3$.
If $1_R+1_R=0$, then $R$ is of characteristic $2$, so $R\cong \mathbb{F}_2$ and $|R|=2$. And, indeed, in this case the hypothesis holds.
If $1_R+1_R\neq 0$, then we get $4\cdot 1_R = 1_R$, hence $3\cdot 1_R=0$, so $R$ is of characteristic $3$, and therefore $R\cong\mathbb{F}_3$ and $|R|=3$. Again, the hypothesis holds for this ring.
In summary, $R$ has either $1$, $2$, or $3$ elements, and is either the trivial ring, $\mathbb{F}_2$, or $\mathbb{F}_3$.