There is Gauss's hypergeometric equation
$[z(1-z)\frac{d^2}{dz^2} + [c - (a+b+1)z]\frac{d}{dz} - ab]u(z) = 0$
Apparently, (see [1], example in 14.2), this equation can be rearranged in the form
$[(z \frac{d}{dz} + a) (z \frac{d}{dz} + b) + (\frac{d}{dz}+c)(\frac{d}{dz})]u = 0$
Consider for $\arg(1-z) < \pi$, c is not integer, for simplicity.
$F(z) = \frac{1}{2\pi i}\int_{-i \infty}^{i \infty}\Gamma(a+s)\Gamma(b+s)\Gamma(-s)\Gamma(c-a-b-s)(1-z)^s ds$
Here the contour divides the "going to the left" and "going to the right" series of poles.
I want to show that $F(z)$ satisfies the equation above.
Step 2. Since this is a regular ODE, the equation has 2 non-proportional solutions, and the standard base is $F(a,b;c;z)$ and $z^{1-c}F(1+a-c,1+b-c; 2-c;z)$
I need to express F(z) as a linear combination of the standard solutions. Some useful hints is in evaluation in [1]14.53. They use Barnes' lemma ( [1] 14.52) , and the corollary from 14.53 that I will write down there.
$\Gamma(-t)(1-z)^t = \frac{1}{2 \pi i} \int_{-i \infty}^{i \infty}\Gamma(s-t)\Gamma(-s)(-z)^s)ds$
But, actually, may be we could try evaluate the values of F(z) and standard solutions at 0 and 1. And then solve the linear equation system without those complex integrals.
If you know some more elementary literature about the question, or a hint, that could make the things easier, please, let me know.
[1] The book of Whittaker, Watson, A course of modern analysis.
UPD: continue thinking about the task.