Yes. Enumerate all arithmetic sets $A_1, A_2, ldots$. Choose $n_kin A_k$ with $|n_k|>2^k$. The set ${n_1, n_2, ldots}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.

Actually you may construct such $S$ for each infinite sequence $A_1, A_2, ldots$ of infinite sets: on $k$-th step choose $n_kin A_kcap S$ and $m_kin A_ksetminus S$ so that $n_1, m_1, n_2, m_2, ldots$ remain distinct.