Question: I thought that $\theta$ must be chosen such that $\cosh(\theta)$ has a range that is equal to the domain of $\frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $\cosh(\theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?

Answer: 
So there are singularities at $\pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-\infty,-1)$.

In fact $x=\cosh(\theta)$ is not valid there. The substitution really being used there is $x=-\cosh(\theta)$, so $dx=-\sinh(\theta) d \theta$, so the integral becomes

$$\int \frac{1}{(\cosh^2(\theta)-1)^{3/2}} (-\sinh(\theta)) d \theta.$$

The catch comes when you rewrite $(\cosh^2(\theta)-1)^{3/2}$ as just $\sinh^3(\theta)$. This is not strictly correct, in fact it is $|\sinh^3(\theta)|$ in general. On $(-\infty,-1)$, this absolute value reduces to $-\sinh^3(\theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,\infty)$ because this minus sign cancels out with the minus sign that came with $dx$.

This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
If you look for a primitive in $(1,+\infty)$,

put $$x=\cosh(\theta)$$

with $$dx=\sinh(\theta) d\theta$$

but if you want a primitive in $(-\infty,-1)$, you should put
$$x=-\cosh(\theta)$$

with $$dx=-\sinh(\theta)d\theta$$