The exact Dyson equation of the lesser Green function $G^<$ reads $$G^< = (1+G^\mathrm{r}\Sigma^\mathrm{r}) g^< (1+\Sigma^\mathrm{a}G^\mathrm{a}) + G^\mathrm{r} \Sigma^< G^\mathrm{a} \qquad(*)$$ with $g$ the undressed Green function. This can be found in, e.g., Eq. (39) of this file. Right below Eq. (39) there, it is stated that the entire first part, i.e., $(...)g^<(...)$, vanishes for steady-state systems if the system was in a noninteracting state in the infinite past. Although this simplification is widely used in the literature, I couldn't find a clear proof for it.
So far, I have noticed the following. From the Dyson equation for retard Green function $G^\mathrm{r}=g^\mathrm{r}(1+\Sigma^\mathrm{r}G^\mathrm{r})$, we immediately find $\Sigma^\mathrm{r}=((g^\mathrm{r})^{-1}{G^\mathrm{r}}-1)(G^\mathrm{r})^{-1}$. Then, for the $(...)g^<$ factor above, we have $(1+G^\mathrm{r}\Sigma^\mathrm{r}) g^< = G^\mathrm{r}(g^\mathrm{r})^{-1} g^< $. Up to this point, everything is exact and general. Then the task seems to be arguing $$(g^\mathrm{r})^{-1} g^< =0 \qquad(**)$$ under certain conditions, which is not clear to me. The reason why this looks to be the aim is strongly hinted by the explicit equations of motion of $g^\mathrm{r}$ and $g^<$ when the corresponding system of Hamiltonian $h(t)$ is noninteracting, i.e., $$[i\hbar\partial_t-h(t)+i0^+]g^r(t,t')=\delta(t-t')\\ [i\hbar\partial_t-h(t)]g^<(t,t')=0.$$ This naively seems to imply (**) always holds, which is certainly not the real case. There must be some condition to formulate it clearly.
