Let $G$ be a group and $H\leq G$. Let $B$ be a sub-semigroup of $H$ such that $B\subseteq B^1(B\setminus B^2)$ where $B^{1}:=B\cup\{1\}$, $B^2:=BB$ (we have $B^2\subseteq B$). Now, put $$ \tau :=\{ A\subseteq G: BA\subseteq A\} \; ,\; \tau' :=\{ HD\cup B^1E: \; D,E\subseteq G \; \&\; E\cap BE=\emptyset\}. $$ We have $\tau'\subseteq \tau$. For if $A=HD\cup B^1E$, for some $D,E\subseteq G$, then $$BA=B(HD)\cup B(B^1E)=HD\cup BE\subseteq HD\cup B^1E=A,$$ (since $B$ is a sub-semigroup of $H$).
Now, my question is whether or not $\tau\subseteq \tau'$ (i.e., if $BA\subseteq A$, then $A=HD\cup B^1E$ for some $D,E\subseteq G$ such that $E\cap BE=\emptyset$).
Note that $G,H,B$, and the empty set are always members of $\tau \cap \tau'$, since $BB\subseteq B$, $B^1(B\setminus B^2)=B$, $B(B\setminus B^2)\cap (B\setminus B^2)=\emptyset$, $HG=G$, and $X\emptyset=\emptyset$, for all $X\subseteq G$.
Also, see https://mathoverflow.net/questions/187125/lower-periodic-subsets-of-groups-and-semigroups
Let $G = H = \mathbf Z^2$ and $B = \{(x,0) : x > 0\}$. Then $\tau$ consists of all subsets $A$ of $\mathbf Z^2$ such that $(x,y) \in A \implies (x+1,y) \in A$, while $\tau'$ consists of the subsets which are either (a) all of $\mathbf Z^2$ or (b) a set of the form $\{(x,y) : x \ge f(y), y \in S\}$. In particular the $x$-axis is in $\tau$ but not $\tau'$.
Possibly you will respond to the above example by requiring that $\langle B \rangle = H$, but that won't help, as the following example shows.
Let $G = H = \mathbf Z^2$ and $B = \{(x,y) : x, y \ge 0, x+y>0\}$. Note $B \setminus B^2 = \{(1,0),(0,1)\}$, so the hypothesis $B = B^1(B\setminus B^2)$ holds. Now $\tau$ consists of all "northeast-closed" subsets of $\mathbf Z^2$, while $\tau'$ consists of all subsets which are either (a) all of $\mathbf Z^2$ or (b) the set of elements to the northeast of some subset $E \subset \mathbf Z^2$, no element of which is northeast of another. Now the upper half plane is in $\tau \setminus \tau'$.