How to prove zero classical classical Green's function contribution

By definition: $$ G^T(t,t') = G^>(t,t')theta(t-t') + G^<(t,t')theta(t'-t), G^{tilde{T}}(t,t') = G^>(t,t')theta(t'-t) + G^<(t,t')theta(t-t'). $$ By summing these two equations we readily obtain: $$ G^T(t,t') + G^{tilde{T}}(t,t') = G^>(t,t') + G^<(t,t'). $$ This equality is thus the consequence of the definitions of the Green's functions involved.

I sketch the derivation of the second identity without delving too much on the notation (apparently defined in Kamenev's review). $$ ilangle left(psi^+(t) -psi^-(t)right)left(psi^+(t') -psi^-(t')right)rangle= ilangle psi^+(t)psi^+(t')rangle + ilangle psi^-(t)psi^-(t')rangle - ilangle psi^+(t)psi^-(t')rangle - ilangle psi^+(t)psi^-(t')rangle= G^T(t,t') + G^{tilde{T}}(t,t') - G^>(t,t') - G^<(t,t') $$ This is zero due to the first equality.

The full set of relations can be found, e.g., in the review by Rammer&Smith.

Update
To add some relevant material that was discussed in the comments. Green's function
$$G(t,t')=-ilangle T_cleft[psi(t)bar{psi}(t')right]rangle,$$ with times $t,t'$ taken on the Keldysh contour, is quivalent to four Green's functions with real time arguments, which are distinguished by whether each time lies on the forward ($C_+/C_1$) or backward ($C_-/C_2$) branch of the Keldysh contour. Specifically, $$G_{12}(t,t') = ilangle psi^+(t)bar{psi}^-(t')rangle,$$ since, according to the countour ordering, $t$ on the forward branch of the contour is "lesser" than $t'$ on the backward branch, $t<_ct'$. The crucial point here is that superscripts $pm$ are only a book-keeping tool for ordering the operators. Once the order is established, they are no more needed, and the above Green's function can be written as $$G_{12}(t,t') = ilangle psi(t)bar{psi}(t')rangle = G^{<}(t,t'),$$ which is by definition the "lesser" Green's function.

Indeed, we can see that the operators taken at the same time $t$ at the two branches of the contour are the same, if we avoid taking the Keldysh limit of extending contour to $+infty$ and stay within the Kadanoff-Baym framework of the contour running from $-infty$ to arbitrarily chosen time and backward to $-infty$. Deforming the contour in such a way that it turns back at time $t$, we see that $psi^+(t)$ is identical with $psi^-(t)$.

Similar manipulations with the contour are also possible when considering two-time Green's function - by deforming the contour one could place the two times either on the same or on the different branches.

Keldysh trick of extending the contour to $+infty$ and seemingly cutting it into two separate branches makes the formalism less transparent, but significantly simplifies the calculations.