= Solution
In <atomic units>, define the <Coulomb integral> and <exchange integral> by
$$
J=\iint\frac{|a(1)|^2|b(2)|^2}{r_{12}}\,d^3r_1d^3r_2,\qquad
K=\iint\frac{a^*(1)b(1)b^*(2)a(2)}{r_{12}}\,d^3r_1d^3r_2.
$$
Expanding the four products in the expectation of $r_{12}^{-1}$ yields
$$
\langle\Phi_\pm|r_{12}^{-1}|\Phi_\pm\rangle=\tfrac12(J+J\pm K\pm K)=J\pm K.
$$
The one-electron terms have the same value for these two states. Within this common-orbital approximation,
$$
E_S=\epsilon_{1s}+\epsilon_{2s}+J+K,\qquad
E_T=\epsilon_{1s}+\epsilon_{2s}+J-K,
$$
so $\boxed{E_S-E_T=2K}$. The <exchange integral> is nonnegative: it is the positive Coulomb quadratic form of the transition density $a^*b$, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital $2K$ an approximate physical splitting.
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