Sets with $\mathrm{d}(h\mathscr{A}) = 1$ but s.t. $h\mathscr{A}$ doesn't contain all large integers

Question

Given $\mathscr{A}\subseteq\mathbb{N}$ an infinite set, consider its $h$-fold sumsets

$$h\mathscr{A}:=\left\{\sum_{i=1}^{h} k_i : k_1,\ldots, k_h\in \mathscr{A} \right\},$$

and let "$\simeq$" be the equivalence relation (in $\mathcal{P}(\mathbb{N})$) given by $\mathscr{A}\simeq \mathscr{B}$ if and only if $\mathscr{A}\triangle \mathscr{B} := (\mathscr{A}\setminus \mathscr{B})\cup (\mathscr{B}\setminus \mathscr{A})$ is finite. Being $\mathrm{d}$ the usual asymptotic density,

Q1. If $h\geqslant 2$, does $\mathrm{d}(h\mathscr{A}) = 1$ imply $h\mathscr{A} \simeq \mathbb{N}$?

This is clearly false for $h=1$, for $\mathscr{A}:= \mathbb{N}\setminus \{n^2: n\in\mathbb{N}\}$ is an easy counterexample. For general $h$ I also think this is false (for if not, the case $h=2$ could be easily used to prove Goldbach's conjecture for suff. large numbers from Vinogradov's results), but I wasn't able to come up with counterexamples. Thus my real question is

How to construct counterexamples to Q1? Furthermore, can we at least guarantee from $\mathrm{d}(h\mathscr{A}) = 1$ that $(h+1)\mathscr{A} \simeq \mathbb{N}$, for all sequences $\mathscr{A}\subseteq \mathbb{N}$?

For the second question, here's a related question of mine in M.SE.

Answer 1

Let $(n_k)$ be a rapidly increasing sequence. Remove from the positive integers all segments $[n_k,2n_k-1]$ and call the remaining set $A$. Then $2n_k$ does not lie in $A+A$, so $A+A$ is not equivalent to $\mathbb{N}$. But $A+A$ has density 1. Indeed, $A$ contains all the numbers from $2n_k$ to $n_{k+1}-1$, thus $A+A$ contains all the number from $4n_k$ to $2n_{k+1}-2$. Also $A+A$ contains $B=2n_k+(A\cap [1,n_k])$ and $C=3n_k+(A\cap [1,n_k])$ which are disjoint sets of cardinalities $n_k+o(n_k)$ each. This implies that if $x\in [2n_k,2n_{k+1}]$ and $k$ is large, then $|(A+A)\cap [1,x]|=x+o(n_k)=x+o(x)$.

For arbitrary $h$ remove the segments $[n_k,hn_k-1]$.