= Solution
\b[No.] Colour the positive integers by alternating dyadic blocks:
$$
C_0=\bigcup_{j\geq0}\big([2^{2j},2^{2j+1})\cap\mathbb N\big),
\qquad
C_1=\bigcup_{j\geq0}\big([2^{2j+1},2^{2j+2})\cap\mathbb N\big).
$$
Every positive integer $x$ and its double $2x$ have opposite colours. A nonempty set closed under addition contains both $x$ and $x+x=2x$, so no such set can lie in one class. In particular there is no infinite monochromatic additive subsemigroup.
The suggested arithmetic-progression argument gives another way to see the obstruction. If $a\in S$ and $S$ is closed under addition, then $a,2a,3a,\ldots$ all lie in $S$. But every infinite <arithmetic progression> of positive step $d$ meets every sufficiently long dyadic block, because a block of at least $d$ consecutive integers contains a representative of the progression's residue class. Both colours contain arbitrarily long blocks, so neither colour contains an infinite <arithmetic progression>.
\Image[/past-exam-of-the-graph-theory-and-probabilistic-combinatorics-course-of-the-university-of-oxford/2010-dyadic-colouring.png]
{title=Alternating dyadic colouring and the doubling obstruction}
{description=Every positive integer and its double lie in opposite coloured blocks.}
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