If E is a subset of a lattice closed under addition then is the intersection of E with the opposite of some translate finite?

this seems intuitive to me but I'm struggling to prove it (is it false?).

Let $E$ be a subset of a lattice (free abelian group of finite rank) closed under addition, containing the origin and such that if $e \in E$ then $-e \notin E$ (it's some sort of "positive" cone, oh actually the correct term should probably be submonoid).

If I take any other element $x$ and consider the intersection of $E$ with $-(E +x)$, will it be finite (or empty)?

(All the properties considered are invariant under isomorphism, so we might as well consider the problem in $\mathbb{Z}^n$)

Since you require $0\in E$ you must exempt $x=0$ from the requirement that $x\in E$ implies $-x\notin E$. If you do, then $E\cap −(E+x)$ may be infinite: take $E=\{(i,j\in\mathbf Z^2\mid i>0 \lor (i=0 \land j\geq0)\}$, and $x=(-n,0)$ with $n\in\mathbf N_{>0}$.