This is an idle curiosity related to Guass's circle problem. Consider any sequence of bounded convex sets $V_1, V_2, \ldots \subseteq \mathbb{R}^2$. Let $L_n$ be the set of lattice points of $V_n$, i.e. $L_n = V_n \cap \mathbb{Z}^2$.
Additionally, suppose we have the following condition (motivated below in Notes): for any $\alpha \in \mathbb{Q}$, for the projection $\pi_\alpha$ from $\mathbb{R}^2$ to the line $y = \alpha x$, $$ \left| \pi_\alpha (L_n) \right| \to \infty. $$ Then must it be the case that $$ \lim_{n \to \infty} \frac{\operatorname{area}(V_n)}{|L_n|} = 1? $$
Notes
If we do not require the condition involving $\pi_\alpha$, then we may take $V_n$ to be a long rectangle along the line $y = \alpha x$ for a counterexample. (Specifically, if the line is $y = 0$ we take the rectangle $[-1.9, 1.9] \times [-n,n]$; if the line is $y = x$ we take the rectangle from $(-n,-n)$ to $(n,n)$ of width almost $\sqrt{2}$, etc.)
I have not specified any condition relating the different convex sets $V_n$; for instance, we don't assume $V_1 \subseteq V_2 \subseteq V_3 \subseteq \ldots$. Thus, I am essentially asking for a sort of uniform convergence of the lattice-point approximation to the area of a convex set, i.e. convergence across all convex sets at once.
I believe my condition is sufficient, but I do not know a proof approach. Perhaps Pick's theorem is useful if we begin by constructing a convex polygon with integer-coordinate vertices around $L_n$.
Del points out in the comments that my conditions do not exclude $V_n$ from being a thin rectangle of some irrational slope. This is true, and requiring the projection condition for $\alpha \in \mathbb{R}$ as well won't help. So maybe this can be used to make a counterexample.