The answer is no, and a counterexample was provided by Turjányi [Tu84]. This was generalised (to the replacement of $A+A$ by the $h$-fold sumset $hA$ for any $h\geq 2$) by Ruzsa and Turjányi [RT85]. Ruzsa and Turjányi do prove (under the same hypotheses) that\[\lim_{N\to \infty}\frac{\lvert (A+A+A)\cap \{1,\ldots,3N\}\rvert}{\lvert A\cap \{1,\ldots,N\}\rvert}=\infty,\]and conjecture that the same should be true with $(A+A)\cap \{1,\ldots,2N\}$ in the numerator.

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