In general, if B contains the origin of $\mathbb{E}$ then, the set erosion and dilation by $B$ become, respectively, anti-extensive and extensive; i.e., for all $X$: $ \begin{equation} X \ominus B \subseteq X \subseteq X \oplus B \end{equation}$
What I'm asking is, if B does not contain the origin of $\mathbb{E}$, is there such a B so that $X \oplus B \supseteq X \ominus B$ ISN'T always true?
The assumption of the erosion and dilation being, respectively, anti-extensive and extensive is only true if and only if the origin is contained in the structuring element.
For example, take a 5x5 zero-matrix with a 1 in the center then we have that dilation and erosion with structuring element,
$$ B = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$
produce the following matrix:
$$ \left[ \begin{array} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] $$
As you see $A \ominus B \not\subseteq A \not\subseteq A\oplus B$.
Edit: Wrong answer but you can solve your question by elemental set theory, if you see that $A \ominus B = \{z / (B_z) \subseteq A \}$ and $ A \oplus B = \{ z / (B_z)\cap A \subseteq A \} $ and take $z\in A \ominus B $ then:
$$ (B_z)\subseteq A \implies (B_z)\cap A \subseteq A \cap A = A$$
Then $z \in A\oplus B$ therefore $A \ominus B \subseteq A \oplus B$.