Let $S_{P Q R S T}=x$. From $S_{\triangle B S T}=S_{\triangle A S R}$, we have $S_{\triangle B T R}=S_{\triangle A T R}$, hence $B A \parallel T R$. Therefore,
$$
\frac{S_{\triangle S T D}}{S_{\triangle S B T}}=\frac{T D}{B T}=\frac{R D}{A R}=\frac{S_{\triangle S R D}}{S_{\triangle S A R}}
$$

Thus,
$$
\begin{array}{l}
S_{\triangle S T D}=S_{\triangle S R D}=\frac{1}{2}(x+1) \\
S_{\triangle A P Q}=S_{\triangle B P Q}=\frac{1}{2}(x+1)
\end{array}
$$

Similarly,
from $S_{\triangle A P Q}=S_{\triangle B P Q}$, we get $B A \parallel P Q$, hence $B A \parallel C E$.
Similarly, $B C \parallel A D, C D \parallel B E, D E \parallel C A, E A \parallel D B$. From $A E \parallel B D$ we have

Thus,
$$
\begin{array}{c}
\frac{S_{\triangle A S D}}{S_{\triangle D S T}}=\frac{A S}{S T}=\frac{E S}{S B}=\frac{S_{\triangle S T E}}{S_{\triangle S B T}} \\
\frac{1+\frac{1}{2}(x+1)}{\frac{1}{2}(x+1)}=\frac{\frac{1}{2}(x+1)}{1}
\end{array}
$$

That is,
$$
\frac{x+3}{x+1}=\frac{x+1}{2}
$$

Thus $x^{2}=5$, so $x=\sqrt{5}$ is the solution. From $A C \parallel D E$ we have
$$
S_{\triangle A R E}=S_{\triangle S R Q}=\frac{1}{2}(\sqrt{5}+1)
$$

Similarly,
$$
S_{\triangle A B S}=S_{\triangle B C T}=S_{\triangle C D P}=S_{\triangle D E Q}=\frac{1}{2}(\sqrt{5}+1)
$$

We can find that $S_{A B C D E}=\frac{1}{2}(15+7 \sqrt{5})$.
Example 7 As shown in Figure 1.6.7, in the convex pentagon $A B C D E$, $A C, A D$ intersect $B E$ at points $S, R$; $C A, C E$ intersect $B D$ at points $T, P$; $C E$ intersects $A D$ at point $Q$. If $S_{\triangle A S R}=S_{\triangle B T T}=S_{\triangle C D T}=$ $S_{\triangle D Q P}=S_{\triangle E R Q}=1$, find $S_{P Q R S T}$.