This answer replacees my earlier, incorrect one. The answer is still "no", even if $x$ does not involve $\alpha$ and $y$ does not involve $\beta$. Example: take $n=2$, $p=3$, $\alpha=\beta=X_1^2X_2^8$, $\epsilon=1$, $x=X_1^4$, $y=-X_2^{16}$. We have $(\alpha+x)(\beta+y)=X_1^6X_2^8-X_1^2X_2^{16}$ which contains no $2\epsilon$-monomials.

This answer replacees my earlier, incorrect one. The answer is still "no", even if $x$ does not involve $\alpha$ and $y$ does not involve $\beta$. Example: take $n=2$, $p=3$, $\alpha=\beta=X_1^2X_2^8$, $\epsilon=1$, $x=X_1^4$, $y=-X_2^{16}$. We have $(\alpha+x)(\beta+y)=X_1^6X_2^8-X_1^2X_2^{24}$ which contains no $2\epsilon$-monomials.

Yes, Rinmyaku modified the question after my counterexample. If $\alpha$ does not appear in $x$ AND $\beta$ does not appear in $y$ then $(\alpha+x)(\beta+y)$ does contain a $2\epsilon$-monomial, namely, $\alpha\beta$.

This answer replacees my earlier, incorrect one. The answer is still "no", even if $x$ does not involve $\alpha$ and $y$ does not involve $\beta$. Example: take $n=2$, $p=3$, $\alpha=\beta=X_1^2X_2^8$, $\epsilon=1$, $x=X_1^4$, $y=-X_2^{16}$. We have $(\alpha+x)(\beta+y)=X_1^6X_2^8-X_1^2X_2^{16}$ which contains no $2\epsilon$-monomials.