$m=r+d, n=1+d \prod\limits_{p<m} p^2$. It is straightforward to check that $|S(m,1)| = r+d$ and (via mobius / PIE) $$S(m,n) = \sum_{k \text{ squarefree }, k\le m} (-1)^{\omega(k)} \lfloor m/k \rfloor\lfloor n/k \rfloor$$ Thus, $$S(m,n)-S(m,1) = \sum_{k \text{ squarefree }, k\le m} (-1)^{\omega(k)} \lfloor m/k \rfloor d \prod\limits_{p\le m} p^2/k$$ This term is divisible by $d$, so we are done.