Nice way of showing the locus of $\left|z-\frac{1}{z}\right| = 2$ is the union of 2 intersecting circles. For visualisation, the following code was run in Mathematica.
ContourPlot[{Abs[x + I y - 1/(x + I y)] == 2}, {x, -2.42, 2.42}, {y, -2.42, 2.42}, Axes -> True]
Here is the locus plot generated by the above code.

I do not see a nice way of showing the locus is what is is apart from brute force algebraic factorisation, which basically requires knowing the form of the locus a priori.
A less cumbersome approach is desired, if such a method exists.
 A: $$\begin{align}
\left|z-{1\over z}\right|=2
&\iff|z-1||z+1|=2|z|\\
&\iff((x-1)^2+y^2)((x+1)^2+y^2)=4(x^2+y^2)\\
&\iff(x^2+y^2+1-2x)(x^2+y^2+1+2x)=4(x^2+y^2)\\
&\iff(x^2+y^2+1)^2-4x^2=4(x^2+y^2)\\
&\iff(x^2+y^2-1)^2-4x^2=0
\end{align}$$
On the other hand, the equation for the locus of two circles of radius $r$, centered at $(x,y)=(\pm c,0)$, is
$$((x-c)^2+y^2-r^2)((x+c)^2+y^2-r^2)=0$$
This can be rewritten as
$$(x^2+y^2+c^2-r^2)^2-4c^2x^2=0$$
All we need is to let $c=1$ and $r=\sqrt2$.
A: We can bring a little geometry into the picture.
First render
$|z+1|^2|z-1|^2=4|z|^2$
by algebraic methods as explained in other answers.  Now draw the quadrilateral whose vertices are $\pm 1, \pm z$.  This is a parallelogram, meaning the sum of its squared diagonal equals the sum of all four squared sides.  Thereby
$4|z|^2+4=2(|z+1|^2+|z-1|^2)$
Eliminating $|z|^2$ between these two equations and collecting all terms to one side then leads to:
$|z+1|^2|z-1|^2-2(|z+1|^2+|z-1|^2)+4=0$
which is factorable!
$\color{blue}{(|z+1|^2-2)(|z-1|^2-2)=0}$
The union of circles then follows.
A: Preliminary:
$|z-1|^2+|z+1|^2\\=(z-1)(\bar z-1)+(z+1)(\bar z+1)\\=|z|^2-2\,\Re(z)+1+|z|^2+2\,\Re(z)+1\\=2|z|^2+2\tag{Eq1}$
With that we can transform our original equation:
$|z-\frac 1z|=2\iff |z^2-1|^2=4|z|^2$ and there is equivalence because moduli are positive numbers.
We can then factorize
$(|z-1|^2-2)\times(|z+1|^2-2)=|z^2-1|-2(\text{Eq1})+4=4|z|^2-2(2|z|^2+2)+4=0$
In the end we conclude: 

$$|z-\frac 1z|=2\iff |z-1|=\sqrt{2}\quad \text{or}\quad|z+1|=\sqrt{2}$$

Which is the union of two circles.
