Yes, because you can multiply all of the existing numbers in $S$ by $2|S|^2$ and extend $S$ with the numbers $1$ through $|S|$. A minimal solution to the extended problem with the $2|A|=|S|$ constraint will then also be a solution to the original problem without such constraint, just by ignoring what happened to the new integers in it.
If someone gives me a polynomial solver for the constrained problem, I can use this reduction to solve the original problem in polynomial time. Thus, since the original problem is known to be NP-hard, the constrained one is NP-hard too.
For example, $\{100,200,300\}$ without constraint reduces to $\{1,2,3,1800,3600,5400\}$ with constraint. The optimal solution to the latter problem is $\{3,1800,3600\},\{1,2,5400\}$, with a difference of $|5403-5403|=0$. Because this difference is strictly less than $3^2$, the original unconstrained $\{100,200,300\}$ has a solution with difference strictly less than 1.