Example 7 If real numbers $a, b, c, d, e$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
a b+a c+a d+a e=-1, \\
b c+b d+b e+b a=-1, \\
c d+c e+c a+c b=-1, \\
d e+d a+d b+d c=-1, \\
e a+e b+e c+e d=-1,
\end{array}\right.
$$

Find all possible values of $a$.
Let $S=a+b+c+d+e$. Then
$$
\begin{array}{l}
a(S-a)=-1, b(S-b)=-1, \\
c(S-c)=-1, d(S-d)=-1, \\
e(S-e)=-1
\end{array}
$$

Given $\Delta=S^{2}+4>0$, we know that the graph of $f(x)=x^{2}-x S-1$ intersects the $x$-axis at two points. Therefore, $a, b, c, d, e$ can have at most two different values.

If $a, b, c, d, e$ are all equal, let's assume they are all equal to $a$, then $S=5a$. Thus, $a(5a-a)=-1$, which has no real roots, contradicting the problem statement.

If $a, b, c, d, e$ have two different values, let's assume they are $a$ and $b$. By symmetry, we only need to consider two cases.
(1) Four $a$s and one $b$, then $S=4a+b$. Hence,
$$
\left\{\begin{array}{l}
a(4a+b-a)=-1 \\
b(4a+b-b)=-1
\end{array}\right.
$$

The system of equations has no real roots.
(2) Three $a$s and two $b$s, then $S=3a+2b$. Hence,
$$
\left\{\begin{array}{l}
a(3a+2b-a)=-1 \\
b(3a+2b-b)=-1
\end{array}\right.
$$

Solving the system, we get $(a, b)=\left( \pm \frac{\sqrt{2}}{2}, \mp \sqrt{2}\right)$.
In summary, $a$ has only four values, which are $\pm \frac{\sqrt{2}}{2}$ and $\pm \sqrt{2}$.