In the case of atoms or molecules in a vacuum, at $0~\mathrm{K}$, and infinitely apart, the "reaction energy" of
$$
\ce{A + B -> A \bond{....}\bond{....}\bond{....} B},
$$
where $\ce{A \bond{....}\bond{....}\bond{....} B}$ is effectively at infinite separation, is
\begin{align*}
E_{\text{rxn}} &= \sum_{i}^{\text{products}} E_{i} - \sum_{j}^{\text{reactants}} E_{j} \\
&= E_{\ce{A \bond{....}\bond{....}\bond{....} B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= E_{\ce{A}} + E_{\ce{B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= 0.
\end{align*}
The reasoning is that energies are extensive, and there is no change in the total energy of the system. The total energy of infinitely separated particles is their sum, as there is no attractive or repulsive interaction between them. They're effectively completely isolated. In that sense, there is no reaction, which is I put reaction energy in scare quotes.
In practice, when you have a finite interaction such as a chemical reaction, the total interaction energy should include electronic, vibrational, rotational, and translational components, of which the latter 3 will grow as temperature increases. Most likely the thermal vibrational contribution is the only significant one.
One missing term that is not dependent on temperature and should be added is the zero-point (vibrational) energy,
$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c}^{\text{normal}\\\text{modes}} \omega_{c},
$$
which is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. The ZPVE or ZPE can alter energy barriers non-linearly, since vibrations are not necessarily localized, and is the most significant non-electronic contribution to gas-phase reaction energies.