Solution:
Observe that
$$\begin{aligned} x y + \frac{1}{x y} - \frac{y}{x} - \frac{x}{y} &= \frac{(x y)^2 + 1 - x^2 - y^2}{x y} \\ &= \frac{\left(x^2 - 1\right)\left(y^2 - 1\right)}{x y} \\ &= \frac{(x + 1)(y + 1)(x - 1)(y - 1)}{x y} \\ &= (x y + x + y + 1) \frac{(x y - x - y + 1)}{x y} \end{aligned}$$
Since $x y + x + y = 1$, the first term will equal $2$. Moreover, dividing both sides of the equation $x y + x + y = 1$ by $x y$, we obtain
$$1 + \frac{1}{y} + \frac{1}{x} = \frac{1}{x y}$$
which is equivalent to
$$1 = \frac{1}{x y} - \frac{1}{y} - \frac{1}{x}$$
Hence, $(x y + x + y + 1) \frac{(x y - x - y + 1)}{x y} = 2 \cdot 2 = 4$.