Let a semigroup satisfy $F(x,x)\approx x$, where $F$ is a binary operation symbol.Let $B$ satisfy $x(yz)\approx (xy)z$ and $xx\approx x$. Does $B$ satisfy $F$ as a hyperidentity?We need only consider 6 binary terms $x,y,xy,yx,xyx,yxy$ How do we see that this gives us 6 equations all of the form $$x^a\approx x ?$$ where $a$ is a natural number. What are these equations? The definition of hyperidentity follows:
Let $\sigma:\{f_i:i\in I\}\to W_\tau(X)$ be a mapping assigning to every $n_i$-ary operation symbol $f_i$ of type $\tau$ an $n_i$-ary term, $\sigma(f_i)$. Any such mapping $\sigma$ will be called a hypersubstitution of type $\tau$.
Here $W_\tau(X)$ is the usual recursive definition of terms:
$x_1,...,x_n$ are $n$-ary terms
if $w_1,...,w_m$ are $n$-ary terms and $m=n_i$ (for some $i\in I$) then $f_i(w_1,...,w_m)$ is an $n$-ary term.
NOW we can think of any hypersubstitution $\sigma$ as mapping the term $f_i(x_1,...,x_{n_i})$ to the term $\sigma(f_i)$. It follows that every hypersubstitution of type $\tau$ induces a mapping $\hat{\sigma}:W_\tau(X)\to W_\tau(X)$ as follows:for any $w\in W_\tau(X)$, the term $\hat{\sigma}[w]$ is defined by
(1) $\hat{\sigma}[x]:=x$ for any variable $x\in X$
(2) $\hat{\sigma}[f_i(w_1,...,w_{n_i})]:=\sigma(f_i)(\hat{\sigma}[w_1],...,\hat{\sigma}[w_{n_i}]).$
The reason for which it is enough to consider the six binary terms $$t_1(x,y) = x, \; t_2(x,y) = y, \; t_3(x,y) = xy, \; t_4(x,y) = yx, \; t_5(x,y) = xyx, \; t_6(x,y) = yxy$$ is that these form the free band on two generators, $\mathbf F$, and since your interested in an equation on two variables, that's all you need to check.
Now, the result follows from hyper-substituting each of the $t_i$ for $F$ in the equation $F(x,y) \approx x$.
This yields $$t_1(x,x) = t_2(x,x) = x,$$ $$t_3(x,x) = t_4(x,x) = x^2 \approx x,$$ $$t_5(x,x) = t_6(x,x) = x^3 \approx x.$$ The identities $x \approx x^2 \approx x^3$ follow from the fact that $\mathbf B$ satisfies $F(x,x) \approx x$, by hypothesis.