Let $R$ be a non-zero commutative ring with identity. The following is well known:
If $x,y \in R$ are idempotents then $x+y-2xy$ is also an idempotent and more than that by defining $x*y = x+y-2xy$, the set of idempotent elemetns becomes a 2-group.
Now Let $T$ be the set of all elements $x$ with $x^3 =x$. Is there any general binary operation on $T$ that makes $T$ a (semi)group ?
Any set $T$ has a binary operation on it that makes the set into a semigroup! Namely $x\ast y=x$, the "choose the left" operator. (It is easy to check that this operation is associative.)
Only slightly less trivially, any set $T$ has a binary operation on it that makes the set into a group! Fix a group $G$ with the same cardinality as $T$, and fix a bijection $G\cong T$. This endows $T$ with a group structure.
That said, let $R$ be a ring and let $T=\{x\in R\ :\ x^3=x\}$. You might wonder whether there exists a binary function $\ast:T\times T\to T$ which makes $T$ into a (semi)group, is symmetric in the two coordinates, and is defined equationally. For the semigroup question, the answer is also trivially yes, by just taking $\ast$ to be multiplication in $R$!
For the group question, assume $x\ast y=f(x,y)$ where $f(a,b)\in \mathbb{Z}[a,b]$. Since $x^3=x$ and $y^3=y$, and $f$ is symmetrtic, we may write
$$f(a,b)=s_0+s_1(a+b)+s_2(a^2b+ab^2)+s_3(a^2+b^2)+s_4ab+s_5a^2b^2,\qquad s_i\in \mathbb{Z}.$$
In the ring $\mathbb{Z}$ the only elements in $T$ are $0,1$. So one of the two is the group identity, hence either $f(1,1)=1$ or $f(0,0)=0$. Thus, the same holds in any other ring.
Case 1: First, assume $f(0,0)=0$. This implies $s_0=0$. Now, since $0$ is the group identity, $f(x,0)=x$. This implies $s_1=1$ and $s_3=0$ (seen by working over the ring $\mathbb{Z}[x]/(x^3=x)$). Next, we must have $f(1,1)=0$ since in $R=\mathbb{Z}$ the set $T=\{0,1\}$ can only have one group structure with $0$ the identity. Hence $s_5=-2-2s_2-s_4$.
Also $f(x,x)^3=f(x,x)$. But $f(x,x)=(2+2s_2)(x-x^2)$. In the ring $\mathbb{Z}[x]/(x^3=x)$, this cubes to itself only if $s_2=-1$. So we now have $f(a,b)=a+b-a^2b-ab^2+tab-ta^2b^2)$. If we compute $f(x,y)^3-f(x,y)$ in $\mathbb{Z}[x,y]/(x^3=x,y^3=y)$, the coefficient on $xy$ is $4t^3-t$, which implies $t=0$.
With $f(a,b)=a+b-a^2b-ab^2$, we find that $f$ does not give an associative operation.
Case 2: Now assume $f(1,1)=1$. We can reduce to case 1 by using the new polynomial $g(a,b)=1-f(1-a,1-b)$.
