SOLVED

Is there a set $A\subset\mathbb{N}$ such that, for all large $N$,
\[\lvert A\cap\{1,\ldots,N\}\rvert \ll N/\log N\]
and such that every large integer can be written as $2^k+a$ for some $k\geq 0$ and $a\in A$?

Lorentz [Lo54] proved there is such a set with, for all large $N$,
\[\lvert A\cap\{1,\ldots,N\}\rvert \ll \frac{\log\log N}{\log N}N\]
The answer is yes, proved by Ruzsa [Ru72]. Ruzsa's construction is ingeniously simple:
\[A = \{ 5^nm : m\geq 1\textrm{ and }5^n\geq C\log m\}+\{0,1\}\]
for some large absolute constant $C>0$. That every large integer is of the form $2^k+a$ for some $a\in A$ is a consequence of the fact that $2$ is a primitive root of $5^n$ for all $n\geq 1$.

In [Ru01] Ruzsa constructs an asymptotically best possible answer to this question (a so-called 'exact additive complement'; that is, there is such a set $A$ with \[\lvert A\cap\{1,\ldots,N\}\rvert \sim \frac{N}{\log_2N}\] as $N\to \infty$.