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.
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|\ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $\operatorname{arcosh}x$ with $|x|<1$.