Solution:
For $a \neq 0, a \neq 1$, the given equality is equivalent to
$$\begin{aligned} & a\left(x_{1} x_{2}+x_{3} x_{4}\right)=x_{1} x_{2} x_{3} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right) \\ & \Longleftrightarrow 2 a-1=(a-1)(a+5) \Longleftrightarrow a^{2}+2 a-4=0 \\ & \Longleftrightarrow a_{1,2}=-1 \pm \sqrt{5} \end{aligned}$$
It is easy to check that for these values of $a$ both equations have real roots. The case $a=0$ is excluded by the condition and $a=1$ implies $x_{4}=0$, whence $x_{1}=0$, which is a contradiction.