System of five equations I'd really appreciate help with this problem, because I'm stuck with it.
$$
\left\{ 
\begin{array}{c}
ab+ac+ad+ae=-1 \\ 
ba+bc+bd+be=-1 \\ 
ca+cb+cd+ce=-1 \\ 
da+db+dc+de=-1 \\ 
ea+eb+ec+ed=-1
\end{array}
\right. 
$$
I've tried substituting $a+b+c+d+e$ with t but it didn't get me far. I've also come up with equations $(a-b)(c+d+e)=0$ or $(b-c)(a+d+e)=0$.

Let $t=a+b+c+d+e$. Then your equations can be transformed to $a(t-a)=b(t-b)=\ldots=e(t-e)=-1$.
From this $0=a(t-a)-b(t-b)=(a-b)(t-a-b)$ and similarly for all other pairs in place of $a,b$  as you found out yourself.
We conclude that $a=b$ or $a+b=t$ (and equally for other pairs).
If  $a=b=c=d=e$, this leads to $4a^2 = -1$, hence one solution is
$$\tag1 a=b=c=d=e=\pm\frac12 i.$$
For the rest we may assume that not all variables are equal, wlog. assume $a\ne b$. Then $a+b=t$. Since we cannot have both $a=c$ and $b=c$ we conclude $a+c=t$ or $b+c=t$, hence $c=a$ or $c=b$. All in all, we see that $a,b,c,d,e$ take only two values. Up to symmetry, we have one of the following cases: $a\ne b=c=d=e$ or $a=b\ne c=d=e$.
In the first case, $t=a+4b$, i.e. $4ab=-1$ and $ab+3b^2=-1$, hence $3b^2=-\frac34$, i.e. $b=\pm\frac{1}2i$ and $a=-\frac1{4b}=b$, contrary to the assumption $b\ne a$.
From now on assume $a=b\ne c=d=e$.
Then $t=2a+3c$, i.e. $a^2+3ac=-1$ and $2ac+2c^2=-1$, hence $c=-\frac{1+a^2}{3a}$ (note that clearly $a\ne0$), and by inserting $\frac{-4 a^4
 - 2 a^2
 + 2}{9 a^2}=-1 $.
From this we find $-4a^4+7a^2+2=0$, hence $a^2=2$ or $a^2=-\frac14$.
Hence we find solutions by letting (using $a^2=2$)
$$\tag 2 \mathrm{Two\ numbers\ are\ }\pm\sqrt 2\mathrm{,\ the\ other\ three\ are\ }\mp\frac{\sqrt2}{2}.$$
On the other hand $a^2=-\frac14$ leads to $c=a$, hence no additional solution.
In total we have thus two solutions from $(1)$ and twenty solutions from $(2)$.
Note that only the solutions of type $(2)$ are real.
