$\square$ Example 13 Draw a line through the point $P(a, b)(a>0, b>0)$ that intersects the $x$-axis and $y$-axis at points $M, N$ respectively. Try to find the range of values for $O M+O N-M N$.
Let $\angle P M O=\theta\left(0-1\right)$ be monotonically decreasing on $\left(-1, \sqrt{\frac{2 a}{b}}-1\right]$ and monotonically increasing on $\left[\sqrt{\frac{2 a}{b}}-1,+\infty\right)$. Considering that $0<t<1$, we discuss in three cases:
(1) When $1<\frac{2 a}{b}<4$, we have $\frac{a}{2}<b<2 a$, thus $0<\sqrt{\frac{2 a}{b}}-1<1$. The function $g(t)$ is monotonically decreasing on $\left(0, \sqrt{\frac{2 a}{b}}-1\right]$ and monotonically increasing on $\left[\sqrt{\frac{2 a}{b}}-1,1\right)$. Therefore, $g\left(\sqrt{\frac{2 a}{b}}-1\right) \leqslant g(t)<\max \{g(0), g(1)\}, g\left(\sqrt{\frac{2 a}{b}}-1\right)=$ $\sqrt{2 a b}$. Thus, the maximum value of $O M+O N-M N$ is $2(a+b)-2 \sqrt{2 a b}$, and the range of $O M+O N-M N$ is $(\min (a, b), 2(a+b)-2 \sqrt{2 a b}]$.
(2) When $2 b \leqslant a$, we have $\sqrt{\frac{2 a}{b}}-1 \geqslant 1$, the function $g(t)$ is monotonically decreasing on $(0,1)$, thus $g(1)<g(t)<g(0)$, i.e., $2 a+b<g(t)<a+2 b$. Therefore, the range of $O M+O N-M N$ is $(b, a)$, and $O M+O N-M N$ has neither a maximum value nor a minimum value.
(3) When $b \geqslant 2 a$, we have $\sqrt{\frac{2 a}{b}}-1 \leqslant 0$, the function $g(t)$ is monotonically increasing on $(0,1)$, thus $g(0)<g(t)<g(1)$, i.e., $a+2 b<g(t)<2 a+b$. Therefore, the range of $O M+O N-M N$ is $(a, b)$, and $O M+O N-M N$ has neither a maximum value nor a minimum value.