Extend $C B$ and $D A$ to intersect at $F$, let $S_{\triangle A B F}=S_{3}$. Since $C E$ bisects $\angle B C D$ and $C E \perp A D$, we know that $\triangle C D F$ is an isosceles triangle, thus $S_{1} = S_{2} + S_{3}$.
Also, because $D E = 2 A E$ and $A B \parallel C D$, we know that $A F = \frac{1}{4} D F$. Therefore,
$$
\frac{S_{3}}{S_{1} + S_{2} + S_{3}} = \frac{S_{3}}{2 S_{1}} = \left(\frac{A F}{D F}\right)^{2} = \frac{1}{16}
$$

Given that $S_{1} = 1$, then $S_{3} = \frac{1}{8}$, hence $S_{2} = \frac{7}{8}$.
Example 8 As shown in Figure 1.4.9, in trapezoid $A B C D$, $A B / / D C, C E$ is the bisector of $\angle B C D$, $C E \perp A D$ at $E, D E=2 A E, C E$ divides the trapezoid into two parts with areas $S_{1}$ and $S_{2}$, if $S_{1}=1$, find $S_{2}$.