Assuming such a partition is possible, then let them be $S_1$, $S_2$, $\cdots$, $S_k$ where $S_i=\{a_i+nd_i\}$, $i=1$, $2$, $\cdots$, $k$. Consequently, for any $\mid z\mid < 1$, it should hold that $$\sum_{n=1}^{\infty}z^n = \sum_{n=1}^{\infty}z^{a_1+nd_1}+\sum_{n=1}^{\infty}z^{a_2+nd_2}+\cdots + \sum_{n=1}^{\infty}z^{a_k+nd_k}$$

or $$\frac{z}{1-z} = \frac{z^{a_1}}{1-z^{d_1}}+\frac{z^{a_2}}{1-z^{d_2}}+\cdots + \frac{z^{a_k}}{1-z^{d_k}}$$

Now, let $z$ approach $e^{i\frac{2\pi}{d_1}}$. The left side is a finite number. However, the right side will become infinite. Hence, this cannot hold which means the answer to the original question is no.