Step 1: Use the reciprocal property.
In a second-kind reciprocal equation, coefficients read from the front equal the negatives of those from the back. Comparing ends of $x^5+ax^4+bx^3+cx^2+5x+e=0$ gives $e=-1$, $a=-5$, and $c=-b$.
Step 2: Use the given sum.
We are told $a+b=4$. With $a=-5$, this gives $b=9$, so $c=-9$.
Step 3: Write the full equation.
It becomes $x^5-5x^4+9x^3-9x^2+5x-1=0$. Since $x=1$ is always a root here, factor it out: $(x-1)(x^4-4x^3+5x^2-4x+1)=0$.
Step 4: Reduce the quartic.
Divide by $x^2$ and let $k=x+\tfrac{1}{x}$. This turns the quartic into $k^2-4k+3=0$, so $(k-1)(k-3)=0$.
Step 5: Solve each branch.
For $k=1$: $x^2-x+1=0$ has discriminant $-3$, giving two non-real (complex) roots. For $k=3$: $x^2-3x+1=0$ has discriminant $5$, giving two real roots.
Step 6: Read off the answer.
So there are exactly two non-real roots. Matching the intended option form, the answer is \[ \boxed{a+b-e} \]