Solution:
The given equation can be rewritten into
$$(r+s)^2 = r s(r s+1)$$
Since $(r+s)^2 \geq 4 r s$ for any $r, s \in \mathbb{R}$, it follows that $r s \geq 3$ for any $r, s > 0$. Using this inequality, equation (1), and the assumption that $r$ and $s$ are positive, we have
$$\begin{aligned} r+s-r s = \sqrt{r s(r s+1)} - r s & = \frac{1}{\sqrt{1+\frac{1}{r s}}+1} \\ & \geq \frac{1}{\sqrt{1+\frac{1}{3}}+1} = -3 + 2 \sqrt{3} \end{aligned}$$
Similarly, we also have
$$r+s+r s \geq 3 + 2 \sqrt{3}$$
We show that these lower bounds can actually be attained. Observe that if $r = s = \sqrt{3}$, then
$$r+s-r s = -3 + 2 \sqrt{3} \quad \text{and} \quad r+s+r s = 3 + 2 \sqrt{3}$$
Therefore, the required minimum values of $r+s-r s$ and $r+s+r s$ are $-3 + 2 \sqrt{3}$ and $3 + 2 \sqrt{3}$, respectively.