I know there's pretty generic algorithm, but I am stuck a bit. The initial problem is:
$$y'' - y = f(x) \quad y'(0) = 0 \quad y(\pi) = 0$$
so I do:
$$\lambda^2-1 =0 \quad \lambda_{1,2} \pm 1 $$
and:
$$y(x) = C_1e^x+C_2e^{-x}$$
Conditions substitution $y'(0) = 0$:
$$y'(x) = C_1e^x-C_2e^{-x}$$
and we get
$$C_1-C_2 = 0$$
therefore we get a system:
$$\begin{cases} C_1 -C_2 = 0 \\ C_1e^{\pi}+C_2e^{-\pi} = 0 \end{cases} \tag{1}$$
EDIT after the answer:
As far as I understand Green's function is to be found in the form of:
$$G(s,x) = \begin{cases}Ay_1(x) \quad 0 \le x \le s \\ By_2(x) \quad s \le x \le \pi\end{cases} \tag{2}$$
And from there we got the system of equations (here we replace $y(x) \to y(s)$):
$$\begin{cases}By_2(s) = Ay_1(s) \\ By'_2(s) =Ay'_1(s)+1\end{cases} \tag{3}$$
That is where I am stuck, I do not know how to proceed to get the $\operatorname{G}(s)$ finally.
You need a left and right solution, the left satisfying the left boundary condition, the right the right.
With some linear equations solving these should turn out as $$ y_1(x)=A\cosh(x)~~\text{ and }~~y_2(x)=B\sinh(\pi-x) $$ These then get combined into a solution to $$ y''-y=\delta(x-s) $$ meaning that $$ y_1(s^-)=y_2(s^+)~~\text{ and }~~y_1'(s^-)+1=y_2'(s^+). $$ This again is simple linear equation solving to find the appropriate $A,B$.