I think the answer is yes. In fact, we show the existence of such a sequence $S$ where $k=2$, i.e. it contains one of every two consecutive integers. Call such a sequence $2$-syndetic.

We sketch roughly how to construct the sequence inductively; suppose by induction we have a $2$-syndetic sequence $S_\ell \subset [1, N_\ell]$, such that $1, N_\ell \in S_\ell$, and every AP of length $\ell$ in $S_\ell$ has common difference at least $f(\ell)$, where $f$ is some increasing function in $\ell$.

Pick randomly offsets $\epsilon_1, \epsilon_2, \ldots, \epsilon_t \in \{0,1\}$ and define $S_{2\ell}$ to be $S_\ell \cup (S_\ell + N + \epsilon_0) \cup (S_\ell + 2N + \epsilon_0 + \epsilon_1) \cup \ldots$. That is, we paste together shifts of $S_\ell$ spaced either $0$ or $1$ apart, where the spacing is picked randomly and independently.

The answer is yes, in fact the sequence $S\subset \mathbb{N}$ indexed by Thue-Morse sequence works, see this MO question. Note that this sequence is syndetic because it contains one of $\{2n, 2n+1\}$ for every $n$. You can extend to the negatives by reflecting and trivially only lose a factor of two on $\ell$.