All (pseudo)varieties considered here are (pseudo)varieties of monoids.
It is known that any (finite or infinite) monoid that satisfies the identities \begin{equation} xhxyty = xhyxty, \quad xhytxy=xhytyx \tag{$*$} \end{equation} is finitely based. Therefore every subvariety of the variety $\mathbf{V}$ defined by $(\ast)$ is finitely based.$^\dagger$ How about the pseudovariety $\mathbb{V}$ defined by $(\ast)$? Is every subpseudovariety of $\mathbb{V}$ finitely based?
$^\dagger$Each subvariety of $\mathbf{V}$ can be defined by $(\ast)$ together with finitely many of the following identities:
(1)\begin{equation} x^{e_0} \prod_{i=1}^r (h_ix^{e_i}) = x^{f_0}\prod_{i=1}^r (h_ix^{f_i}), \end{equation} where $e_0,f_0,\ldots,e_r,f_r \geq 0$ and $r \geq 0;$
(2)\begin{equation} x^{e_0} y^{f_0} \prod_{i=1}^r (h_ix^{e_i}y^{f_i}) = y^{f_0} x^{e_0} \prod_{i=1}^r (h_ix^{e_i}y^{f_i}), \end{equation} where $\ \ e_0,f_0 \geq 1,\ $ $e_1,f_1,\ldots,e_r,f_r \geq 0,\ $ $\sum_{i=0}^r e_i \geq 2,\ $ $\sum_{i=0}^r f_i \geq 2,\ $ and $\ \ r \geq 0.$