The figure on which E_0 and E_A are plotted, compares the situation of tm0 with that of tmA. But these can actually not be compared in such a straightforward way. The electron-charge distributions do not have the same total charge: tm0 has -2e and tmA has -2e-\epsilon. If you would distribute the charge -(2e+\epsilon) over two point charges (resulting in the situation like tm0), then you would have a lower energy E’_0 than what is currently shown in the plot of E_0. It’s with E’_0 that tmA should be compared to.