First part. Assume that the equations (2) have real roots satisfying $x_1y_1 - x_2y_2 = 1$. By the familiar formula, the roots of the quadratic equations are given by
$$x_{1,2} = \frac{-p \pm K}{2} \quad \text{and} \quad y_{1,2} = \frac{p \pm L}{2},$$
where the real numbers $K, L$ satisfy $K^2 = p^2 - 4q$ and $L^2 = p^2 - 4r$ (we choose the signs of $K, L$ in accordance with the labelling of the roots). Thus
$$1 = x_1y_1 - x_2y_2 = \frac{(-p+K)(p+L) - (-p-K)(p-L)}{4} = \frac{p(K-L)}{2},$$
whence $p \neq 0$ and $K - L = 2/p$. Substituting this into the equality
$$(K+L)(K-L) = K^2 - L^2 = (p^2-4q) - (p^2-4r) = 4(r-q)$$
yields $K+L = 2p(r-q)$. From these values of $K+L$ and $K-L$ we obtain $K = 1/p - p(q-r)$, so upon squaring, $K^2 = 1/p^2 - 2(q-r)p^2 + (q-r)^2$. Comparing this with the equality $K^2 = p^2 - 4q$, an easy manipulation leads to the desired equation (1).

Second part. Assume that (1) holds. Then clearly $p \neq 0$. The equation (1) can be rewritten in either of the following two forms,
$$p^4(q-r)^2 + 2p^2(q-r) + 1 = p^4 - 4p^2r \quad \text{and} \quad p^4(r-q)^2 + 2p^2(r-q) + 1 = p^4 - 4p^2q.$$
Upon dividing by $p^2$ we find that the discriminants of the equations (2) are equal to
$$p^2 - 4q = \left( \frac{p^2(r-q)+1}{p} \right)^2 \quad \text{and} \quad p^2 - 4r = \left( \frac{p^2(q-r)+1}{p} \right)^2;$$
hence, they are nonnegative and the (real) roots of (2) have the form (3), where
$$K = \frac{p^2(r-q)+1}{p} \quad \text{and} \quad L = -\frac{p^2(q-r)+1}{p}.$$
The signs of the numbers $K$ and $L$ have been chosen so that (see First Part)
$$x_1y_1 - x_2y_2 = \frac{p(K-L)}{2} = \frac{p}{2} \cdot \left( \frac{p^2(r-q)+1}{p} + \frac{p^2(q-r)+1}{p} \right) = 1.$$