Slightly explained: Let $a b=c d=t(t<0)$. Then
$$a<0<b, c<0<d \text {. }$$

Construct the functions
$$\begin{array}{l} f(x)=x^{2}-(a+b) x+t, \\ g(x)=x^{2}-(c+d) x+t . \end{array}$$

Then $a, b$ are the two roots of the equation $f(x)=0$, and $c, d$ are the two roots of the equation $g(x)=0$.
$$\begin{array}{l} \text { Also, } f(c)=c^{2}-(a+b) c+c d \\ =c[(c+d)-(a+b)]<0, \end{array}$$

Then $a<c<d$.
Similarly, $c<b<d$.
Therefore, $a<c<b<d$.
$$\text { Hence } M \oplus N=(a, c) \cup(b, d) \text {. }$$