Let $B, E$ be nonempty subsets of a group or semigroup $X$ such that $E\cap BE=\emptyset$, and put $A=BE\cup E$. Then
(a) if $B$ is a sub-semigroup of $X$, then $BA\subset A$ (i.e., $BA$ is a proper subset of $A$). Because $BA=B^2E\cup BE$, $B^2\subseteq B$ (where $B^2:=BB$) and so $BA=BE=A\setminus E\subset A$.
(b) is there any example such that $BA\subset A$ but $B$ is not a sub-semigroup of $X$? (to show that the converse is not true)
Note that a sufficient condition for $BA\subset A$ is $B^2E\subseteq BE$ (e.g., if $B$ is a sub-semigroup of $X$, but this condition does not seem necessary). Also, $B^2E\subseteq BE$ is a necessary condition for $BA\subset A$ if $E\cap B^2E=\emptyset$.
A simple example is $X=(\mathbb{Z},+)$, $E=2\mathbb{N}$, and $B=\{1\}$.