Summary: I consider the limit $\lim_{n\to\infty}\prod_{i=1}^n i^{\chi(i)}$ (let me drop the $\mod s\mathbb Z$ for brevity). If we restrict to $n$ divisible by $s$, then the limit will always be equal to zero. If we consider the limit over all $n$, the limit will never exist. [The question was edited to only ask about the first of those cases since I've started writing this, so parts of this answer are now irrelevant. For interest I've decided to keep them though, they are enclosed in square brackets]
[The limit you claim is zero, $\lim_{n\to\infty}\prod_{i=1}^n i^{\chi(i)}$ with $\chi:\mathbb Z/2\mathbb Z\to\{\pm 1\}$, actually does not exist. It only exists and is equal to $0$ if you group the terms into pairs, which amounts to only looking at $n$ even. If you restrict to $n$ odd, then the product will tend to infinity by a very similar argument.]
For the products with arbitrary $s$, let me first consider the sum up to $n=ms$ for some $m$. The logarithm of your product is $-\sum_{k=1}^\infty\chi(k)\log k$. Splitting into residue classes modulo $s$, we have for $a=1,\dots,s$ a sum $\chi(a)\sum_{i=0}^{m-1}\log(is+a)$. Finding asymptotics for those sums can be done using the Stirling formula for the gamma function:
$$\prod_{i=0}^{m-1}(is+a)=s^m\prod_{i=0}^m(\frac{a}{s}+i)=s^m\frac{\Gamma(\frac{a}{s}+m)}{\Gamma(\frac{a}{s})},$$
so by Stirling the logarithm is
$$\sum_{i=0}^{m-1}\log(is+a)=m\log s+\log\Gamma(\frac{a}{s}+m)-\log\Gamma(\frac{a}{s})\\
=m\log s+(\frac{a}{s}+m)\log(\frac{a}{s}+m)-(\frac{a}{s}+m)-\frac{1}{2}\log(\frac{a}{s}+m)+O(1)\\
=m\log(\frac{a}{s}+m)+m(\log s-1)+(\frac{a}{s}-\frac{1}{2})\log(\frac{a}{s}+m)+O(1).$$
Note we have $\log(\frac{a}{s}+m)=\log m+\log(1+\frac{a}{sm})=\log m+O(\frac{1}{m})$, so this sum is further equal to
$$m\log m+m(\log s-1)+(\frac{a}{s}-\frac{1}{2})\log(m)+O(1).$$
Now multiply by $\chi(a)$ and sum over all $a$. Since $\sum_{a=1}^s\chi(a)=0$, all terms not involving $a$ cancel out and we are left with
$$\sum_{n=1}^{ms}\chi(n)\log n=\frac{1}{s}\log m\cdot\sum_{a=1}^s a\chi(a)+O(1).$$
Therefore the behavior of your product depends on what the sum $\sum_{a=1}^s a\chi(a)$. If this sum turns out to be a complex number of positive real part [which it does in the case of interest, see below], then the product will tend to zero. For $\chi$ as at the start this is indeed what happens: $1\chi(1)+2\chi(2)=-1+2=1$. If this turned out to have negative real part, then the sum would tend to infinity. If the real part of the sum is zero, then the exact behavior will depend on the $O(1)$ term. We could very well make that computation too, it would be similar but more messy.
[Now, what happens if we take $n$ to not be divisible by $n$, but instead say of the form $n=ms+c$ for some $c$? In that case, the easiest thing to do given computations already done above is separate some of the original terms:
$$\prod_{i=1}^n i^{\chi(i)}=\prod_{i=1}^c i^{\chi(i)}\prod_{i=c+1}^{ms+c}i^{\chi(i)}.$$
The first factor is a nonzero constant, so doesn't impact the convergence properties. The same calculation as above will establish the same kind of result as for $c=0$, except it will instead depend on the sum $\sum_{a=c+1}^{c+s} a\chi(a)$. For your original $\chi$ and $c=1$ this gives $2\chi(2)+3\chi(3)=2-3=-1$, which is indeed what we observe with the limit over odd $n$ being infinite.]
All that said, the behavior of the sum revolves around sums of the form $\sum_{a=c+1}^{c+s} a\chi(a)$. Let me indicate how to find (and study) them. First consider the case of $c=0$ [which after the edit is the only relevant one]. Let $z=\chi(1)$. Since $\chi$ is a character, $\chi(a)=z^a$, and so our sum becomes $\sum_{a=1}^sa z^a$. This sum is equal to
$$\frac{z(sz^{s+1}-(s+1)z^s+1)}{(z-1)^2}=\frac{z(sz-(s+1)+1)}{(z-1)^2}=\frac{sz(z-1)}{(z-1)^2}=\frac{sz}{z-1}$$
since $z^s=\chi(s)=1$ and $z\neq 1$ since $\chi$ is nontrivial. This value will always have positive real part - there is a simple geometric argument: the argument of this value is the angle formed by points $z,0,z-1$. But since $|z-0|=|z-(z-1)|=1$, the triangle they form will be isosceles, so the angle has to be acute. Therefore, in light of all that was said above, the product over $n$ divisible by $s$ will always tend to zero.
[If we permit other values of $c$, then we can just make the transformation $$\sum_{a=c+1}^{c+s} a\chi(a)=\sum_{a=1}^s(a+c)\chi(a+c)=\chi(c)\sum_{a=1}^sa\chi(a)+c\chi(c)\sum_{a=1}^s\chi(a)=\chi(c)\sum_{a=1}^sa\chi(a)=\frac{sz^{c+1}}{z-1}.$$
This sum can now have real part positive, negative, or zero, so its behavior can vary. For $c=s-1$ this term will become $\frac{s}{z-1}$ and will always have negative real part (since $z-1$ does), so the product will tend to infinity. In particular we get that the product will never exist if we take it over all values of $n$]
