(technically, we should probably write [x],[w] and [x][w] = [xw] = [1],
to avoid confusion with actual integers, but this is rarely done in practice).
we also have z in Zn, with wz = 1.
so we need to find some u in Zn so that (xw)u = 1.
the obvious candidate is zy, so let's see if that works:
(xw)(zy) = x(wz)y = x(1)y = xy = 1.
but let's look at this "on the integer level":
[x][y] = [1] means [xy] = [1], that is: xy = 1 + kn (for some integer k).
[w][z] = [1] means wz = 1 + rn, for some integer r.
now we want to show that there is some u such that [xw][u] = [1].
the u we choose is u = zy.
now [xw][zy] = [(xw)(zy)] = [x(wz)y].
remember, wz = 1 + rn, so
x(wz)y = x(1 + rn)y = (xy)(1 + rn)
(since ordinary multiplication of integers is commutative)
= (1 + kn)(1 + rn) = 1 + (k + r + krn)n,
which shows that [xw][u] = [xw][zy] = [(xw)(zy)] = [1 + (k + r + krn)n] = [1],
as desired.