$\tau(n) \le k$ implies that $n=\prod_{i=1}^j p_i$ with $j\le k$, thus $$\sum_{n=1,\tau(n)\le k}^\infty n^{-s} \le (1+\sum_{p \ prime} p^{-s})^k=\sum_n c_k(n) n^{-s}$$

(a coefficient-wise bound) 

$\pi(x)=O(x/\log x)$ implies that $\sum_{n\le x} c_n(k) =o(x)$ thus $$\sum_{n\le x,\tau(n)\le k} 1=o(x)$$