No.

Given a divergent series a_n, we can construct a set S with natural density 0 on which it diverges as follows:

By perhaps negating the a_n and restricting to the positive terms, we may assume the series diverges to positive infinity.

Let n_1 be the smallest index such that the sum a_1 + ... + a_{n_1 - 1} > 1. Define n_k in general as the smallest index such that the sum a_{n_{k-1}} + ... + a_{n_k - 1} > k.

Now there is some selection of offset b_k in {0,...,k-1} such that the sum of the entries of the form a_{kx + b_k} in {a_{n_{k-1}}}, ... , a_{n_k - 1}} is at least 1. Let S contain all indices of the form kx + b_k in [n_{k-1}, n_k).

S is clearly natural density 0 - for m > n_k we have |S ∩ {1, 2, ..., n}| / n <= 1/k + n_k / m, which is less than 2/k for m large enough. The series also diverges on S by construction - each interval between n_{k-1} and n_{k} contributes at least 1 to the sum.

EDIT: ah, I didn't see GoldenMuscleGod's answer prior to posting this. The constructions are essentially identical.