Well, let's say beggining set A has just elements 1 and x. So, 1 and x are relatively prime and we have new element x+1. That goes in infinity, so we have prove that A coincidences with set of positive integers.

We note that if we have two consecutive integers $m$ and $m+1$ in the set, we have $\frac{m+2}{(m+1, m+2)} = m+2$ in the set. So if we can show that we have $2$, we are done. Suppose we have $1$ and $k$ in the set. If $k$ is odd, choosing $m=1$ shows $2\in A$ If $k$ is even, choose $m=k$, and now we have $k+1$, an odd number, in the set. Set $m=1$ again, and we are done.