We claim the answer is no. Note that for the two integers to be consecutive, their difference must be equal to $1$. Each move maps the set $(m,n) \rightarrow (m,n+m)$ or $(m,n) \rightarrow (m+n,n)$. Thus the difference between two elements of the set in iteration $k+1$ must have been an element that occurred in iteration $k$. So for the numbers to be consecutive, $1$ must occur in the set. However, this is not possible as there is no way to reduce the difference and the smallest number that occurred is $2$, so we are done. $\blacksquare$.