PARTITION with 0-sum assumption
No. The problem is trivial, since taking $I$ to be the empty set always results in a solution to the problem.

Yes, it is NP-complete. Note that if there is a solution, that is

$\Sigma_{i\in I} x_i = \Sigma_{i\not\in I} x_i$

then

$\Sigma_{i\in I} x_i - \Sigma_{i\not\in I} x_i = 0$.

Since we have

$\Sigma_{i\in I} x_i + \Sigma_{i\not\in I} x_i = 0$

then $\Sigma_{i\not\in I} x_i = 0$ and $\Sigma_{i\in I} x_i = 0$.

Proof of NP-hardness is by reduction from partition. For a given instance of the partition problem, a multiset $S$ of integers, where $\Sigma_{s\in S}=2N$, create an instance of Partition with 0-SUM, by adding 2 copies of $-N$ to $S$.

Clearly, if the Partition instance has a solution then so does the modified problem, by adding $-N$ to each half of the partition. If the Partition with 0-SUM restriction has a solution, then so does the original Partition problem: since all the integers in the original partition problem are positive, each 0-SUM "half" of the 0-SUM partition must contain one of $-N$, and the rest must sum to $N$.