No. There are countably many arithmetic progressions $\{n,n+k,\dots\}$; enumerate them in some order. Then recursively prescribe elements of $A$ and $B$: given the $j$th arithmetic progression, insist that two large integers (larger than anything in $A$ or $B$ already) whose sum lies in the AP must be in $A$, and two other large integers whose sum lies in the AP must be in $B$. The result will be disjoint infinite subsets $A_0$ and $B_0$; set $A=A_0$ and $B=\Bbb N\setminus A$, and neither $2A$ nor $2B$ will contain any full arithmetic progression.

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.