Shapness in a theorem of Schnirelmann

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A famous theorem due to Schnirelmann says that:

Suppose that $A,B \subset \mathbb{N}_0$, $0 \in A,B$ and $\sigma A + \sigma B \geq 1$. Then $A + B = \mathbb{N}_0$.

where $\sigma X := \inf_{n} \frac{|X \cap [1,n]|}{n}$, and $X + Y = \{x + y \ : \ x + y\}$. In particular, if $A \subset \mathbb{N}_0$, and $0 \in A$, $\sigma A \geq \frac{1}{2}$, then $A + A = \mathbb{N}_0$.

I am interested in cases when the above observation is nearly sharp. In other words, I would like a set $A$ with $0 \in A$ and $\sigma A \geq \frac{1}{2} - \epsilon$, with $\epsilon$ as small as I please, such that $A+A$ is not all of $\mathbb{N}_0$, and preferably $\mathbb{N} \setminus (A+A)$ is as large as possible. This can be achieved by either of the following constructions.

Construction 1: Pick a large odd integer $q$, and put $A = \{n : n \bmod{q} \in [0,\frac{q-3}{2}] \}$. Then $A$ will have density $\frac{1}{2} - O(\frac{1}{q})$, while it will fail to contain any integer congruent to $-1 \bmod {q}$.

Construction 2: Pick some rapidly increasing sequence $N_k$, such as $N_k = N_0 2^{k^2}$. Define $A = \mathbb{N}_0 \setminus \bigcup_{k=0}^\infty [N_k,2N_k]$. Then, $2N_k \not \in A$, while $\sigma A \geq \frac{1}{2} - O(\frac{1}{N_0})$.

However, I am not quite satisfied with these constructions. Construction 1 just exploints sharpness in Cauchy-Davenport, and lifts the example from $\mathbb{Z}/q\mathbb{Z}$ to $\mathbb{N}_0$. In particular, the resulting set $A$ is very structured. Construction 2 can be generalised somewhat, but the main idea is just to take a sequence $N_k$, and brute-force $2N_k \not \in A+A$ while greedily including as many elements as possible. As a result, the complement of $A+A$ is rather sparse.

What I am really looking for are interesting examples when (the consequence of) the mentioned theorem is nearly sharp. To make it slightly less vague, the set $A$ I am looking for should satisfy, apart from $\sigma A \simeq \frac{1}{2}$ and $0 \in A$, the following conditions:

  1. For any $q$, we have $A+A \bmod q = \mathbb{Z}/q\mathbb{Z}$ (it's even better is $A$ is equidistributed in classes modulo $q$),
  2. The complement of $A+A$ is large, e.g. $|[1,N] \setminus (A+A)| \gg \log N$.