According to Kuratowski's closure-complement problem, the monoid generated by the complement operator $a$ and the closure operator $b$ has $14$ elements and is presented by the relations $a^2 = 1$, $b^2 = b$ and $(ba)^3b = bab$. Now you are interested by the submonoid generated by the **closure operator** $b$ and by the **interior operator** $i = aba$. This submonoid has only $7$ elements: $1$, $b$, $i$, $bi$, $ib$, $bib$ and $ibi$. You can use Kuratowski's example $$K = {]0,1[} \cup {]1,2[} \cup \\{3\\} \cup ([4,5] \cap \mathbb{Q})$$ to generate the $14$ sets and hence the $7$ sets you are interested in. This is an example in $\mathbb{R}$, but $K \times \mathbb{R}$ should work for $\mathbb{R}^2$.