You only need to choose a branch cut so that all of the branch points lie on the cut. Remember that a branch point is a point where the function is discontinuous when traversing an arbitrarily small circuit about the point. By choosing a branch cut like this you're essentially acknowledging this behavior and choosing specifically where this discontinuity will occur.
The points $z = \pm 1$ are clearly branch points, and they are the only finite branch points. The point at infinity may also be a branch point. In this case it isn't (shown below), but if it were then you would need to choose your cut so that it contains all three branch points ($1$, $-1$, and $\infty$). You could do this by choosing it to be the real interval $(-\infty,1]$, but there are infinitely many other choices you could make too.
To see that the point at infinity is not a branch point, rewrite the function as
$$
f(z) = z\,\left(\frac{1}{z}-1\right)^{3/5}\left(\frac{1}{z}+1\right)^{2/5}.
$$
Let $|z| > 1$. By doing this, we can ensure that neither of the two quantities $a = \frac{1}{z}-1$ or $b = \frac{1}{z}+1$ will make a circuit about the origin, regardless of what $z$ is doing. They are both restricted to unit disks centered at $-1$ and $1$, respectively. Indeed, $|a+1|<1$ and $|b-1|<1$.
Now, with this condition on $|z|$, if $z$ makes a circuit about the origin (equivalently, a circuit about infinity), the change in the argument of $f$ is precisely $2\pi$, and the function remains unchanged. We have thus continuously traversed an arbitrarily small circuit about infinity, and hence infinity is not a branch point of $f$.
As a result, we only need to choose a branch cut which includes $1$ and $-1$. The real inverval $[-1,1]$ is the simplest choice.
If $\phi(z)$ is the branch of $\sqrt{z^2-1}$ with a branch cut at $(-\infty,1]$ which is positive on $(1,\infty)$, then $\phi(2i)=\sqrt{5}i$ and $\phi(-2i)=-\sqrt{5}i$. So (writing $\text{Log}$ for the principal value of the logarithm as Gamelin does)
$$\begin{align*}
\text{Log}\left(z+\phi(z)\right)\bigg|_{-2i}^{2i}
&=\text{Log}\left(2i+\phi(2i)\right)-\text{Log}\left(-2i+\phi(-2i)\right)\\
&=\text{Log}((2+\sqrt{5})i)-\text{Log}((2+\sqrt{5})(-i))\\
&=\pi i,
\end{align*}$$
in agreement with the numerical result.
To see that $\phi(2i)=\sqrt{5}i$, let $z$ move along a path from $2$ to $2i$ in the closed first quadrant, for example, $\theta\mapsto z(\theta):= 2e^{i\theta}$, $0\le \theta\le \pi/2$. Then $z(\theta)^2-1=4 e^{2i\theta}-1$ will travel from $3$ to $-5$, always remaining in the closed upper half-plane. Since $\phi(z)$ was defined to be positive on $(1,\infty)$, $\phi(2)=\sqrt{3}>0$, so $\phi(z)$ must also travel along a path in the closed first quadrant. Therefore $\phi(2i)$ is $\sqrt{5}i$ rather than $-\sqrt{5}i$. Proving that $\phi(-2i)=-\sqrt{5}i$ can be done in the same way, by moving $z$ on a path from $2$ to $-2i$ in the closed fourth quadrant and observing that, since $z^2-1$ moves along a path in the closed lower half-plane, $\phi(z)$ must travel along a path in the closed fourth quadrant.
Best Answer
If we are taking $\mathrm{Log}$ on the principal branch as you say, then we know that $\mathrm{Arg}(\mathrm{Log}(z)) \in (-\pi, \pi]$. So the reason that the proof fails is because you can have, for example, $\mathrm{Arg}(\mathrm{Log}(z)) = \mathrm{Arg}(\mathrm{Log}(w)) = \pi$, so you then get
$$\begin{eqnarray} \mathrm{Arg}(\mathrm{Log}(zw)) & = & 0 \\ & \neq & 2 \pi & = & \mathrm{Arg}(\mathrm{Log}(z)) + \mathrm{Arg}(\mathrm{Log}(w)) \end{eqnarray}$$