No. As suggested by S. Carnahan (with the exact numbers tweaked),
$$
A = \bigcup_{k=0}^\infty [3^{2k},3^{2k+1}) \quad\text{and}\quad B = \Bbb N\setminus A = \bigcup_{k=0}^\infty [3^{2k+1},3^{2k+2}).
$$
Then
$$
2A \subset \bigcup_{k=0}^\infty [3^{2k},2\cdot3^{2k+1}) \quad\text{and}\quad 2B \subset \bigcup_{k=0}^\infty [3^{2k+1},2\cdot3^{2k+2}),
$$
and hence both $2A$ and $2B$ have arbitrarily large gaps; this prohibits either set from containing an infinite arithmetic progression.