In the proof of Proposition 2.5 in Silverman's Arithmetic of Elliptic Curves, the author defines a map $$E_{ns} \to \overline{K}^*, \quad [X,Y,Z] \mapsto 1 + \frac{AX}{Y},$$ where $E_{ns}$ is the nonsingular part of the curve $$E : Y^2 Z + A XYZ - X^3 = 0.$$ He then makes the coordinate change $$X = A^2 X' - A^2 Y', \quad Y = A^3 Y', \quad Z = Z'$$ (and divides by an $A^6$ and drops the primes) to get $$E : XYZ - (X - Y)^3 = 0.$$ The author then dehomogenizes by setting $Y = 1$, giving $E : xz - (x - 1)^3 = 0$, where $x = X / Y$ and $z = Z / Y$, and says that the map above is now given by $$E_{ns} \to \overline{K}^*, \quad (x,z) \mapsto x.$$ I don't see how the map in this new coordinate system is given by explicitly by what is written above.
My attempt to find the map was to start in the coordinate system given by $X', Y', Z'$, then apply the inverse of the original change of coordinates to get back to the system $X, Y, Z$, then apply the original map. My thought is that the composition should be how to write the map explicitly in the new coordinates, but when I do that, I get $[X', Y', Z'] \mapsto (1 + A) + A^2 \frac{X'}{Y'}$, which would be $(x,z) \mapsto (1 + A) + A^2 x$ in the dehomogenized coordinates.
Edit: On the off-chance someone reads this in the future, the mistake I made above is thinking of the coordinate change in the wrong direction. The coordinate transformation from $[X',Y',Z']$ to $[X,Y,Z]$ is $[X',Y',Z'] \mapsto [A^2 X' - A^2 Y', A^3 Y', Z']$. So the accepted answer is in fact what you get when you compose the transformation with the original map to $\overline{K}^*$.
It's clearer when you write the new equation as $$E : X'Y'Z' - (X' - Y')^3 = 0.$$
The map in the old coordinates is given by $$[X,Y,Z] \mapsto 1+ \frac{AX}{Y}.$$ And in the new coordinates, you just need to replace $X, Y, Z$ by their values so $$[X',Y',Z'] \mapsto 1 + \frac{AX([X',Y',Z'])}{AY([X',Y',Z'])} \mapsto 1+ \frac{A(A^2 X' -A^2 Y' )}{A^3 Y'} = 1+ \frac{X'}{Y'}- 1 = \frac{X'}{Y'}.$$
When you dehomogenize $x = X'/Y'.$