I'm not sure whether $\sum\limits_{x = -\infty}^{\infty}x$ is equal to $0$ or undefined. For example,
$$\sum_{x = -\infty}^{\infty}x = \displaystyle \sum_{x = -\infty}^{-1}x + \displaystyle \sum_{x = 1}^{\infty}x = -\infty + \infty$$
So with that approach it is undefined. However, clearly all the negative elements in the summation cancel with the positive elements, so that makes it seem like it should be zero. So which is it?
On
The rearanging of the coefficitiens is not allowed. To ensure the rearrangement you need to know about the absolute convegence of the series which obviously does not hold.
A typical example is the nonconvergent series $$\sum_{n\in \mathbf N} (-1)^n.$$
The Riemann series theorem might be interesting for you in this context.
On
The answer really depends entirely on what you mean by "the sum $\sum_{n=-\infty}^{n=\infty}x$".
The standard definition for the sum of an infinite series given in most books only holds when the sum is done from $n=0$ to $n=\infty$. (A "one-tailed" series). However, your series is two-tailed and the definition doesn't apply.
So you need to extend your definition first, and that alone will dictate what your answer is.
That being said, a common situation where two-tailed series comes up is in Laurant series (compare with Taylor series). In this case, a Laurant series will have an annulus of convergence (compare to the radius of convergence). Inside the annulus of convergence, the natural way to define the sum is the naive way: add up partial sums starting at $0$ and going out, interleaving positive and negative values of $n$. The sum will converge to a finite value and rearrangement is immaterial.
You can see that the sum does not converge because the limit $$\lim_{(m, n) \to (-\infty, \infty)} \sum_{x=m} ^{n} x\tag{1}$$ does not exist. On the other hand note that the limit $$\lim_{n\to\infty} \sum_{x=-n} ^{n} x\tag{2}$$ exists and is equal to $0$. The sum in your question is defined via $(1)$ and not via $(2)$.