While I realise your question is in the second quantized picture and this may not be the answer you're looking for, I just wanted to offer a simple argument in the first quantized picture that would demonstrate the symmetry of PHS Hamiltonians about $E = 0$ (half filling).
Define $\hat{\mathcal{P}}$ as the particle-hole operator in the second quantized picture. The action of $\hat{\mathcal{P}}$ is defined as
\begin{equation} \hat{\mathcal{P}}\hat{c}^{}{A}\hat{\mathcal{P}}^{-1} = \sum{B} \hat{c}{B}^{\dagger}(U{P})_{B, A} \end{equation}
for some unitary matrix $U_{P}$ (i.e. we turn an electron into a hole and vice-versa). 526
$\mathcal{P}$ is also linear in the sense $\mathcal{P}i\mathcal{P}^{-1} = i$ The action of $\mathcal{P}$ on the Hamiltonian, $H$, is therefore
\begin{equation} \begin{split} \hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} &= \sum_{A, B} \hat{\mathcal{P}} \hat{c}{A}^{\dagger} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} H^{}{A, B} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} \hat{c}^{}{B} \hat{\mathcal{P}}^{-1}\ & =\sum{A, B} \sum_{C, D} (U_{P}^{})^{\dagger}{A, C} \hat{c}{C} H_{A, B} \hat{c}{D}^{\dagger} (U{P}^{})^{}{D, B} \ &= \sum{A, B}\sum_{C, D} \delta^{}{C, D}(U{P})^{\dagger}{A, C}H^{}{A, B}(U_{P})^{}{D, B} - \hat{c}^{\dagger}{D}(U_{P})^{}{D, B}H^{}{A, B}(U_{P})^{\dagger}{A, C}\hat{c}^{}{C} \ & = \sum_{A, B}\sum_{C, D} (U_{P})^{\dagger}{A, C}H^{}{A, B}(U_{P})^{}{C, B} - \hat{c}^{\dagger}{D}(U_{P})^{}{D, B}H^{T}{B, A}(U_{P})^{\dagger}{A, C}\hat{c}^{}{C} \ & = \sum_{A, B}\left( \delta^{}{A, B}H{A, B} - \sum_{C, D}\hat{c}^{\dagger}{D}(U{P})^{}{D, B}H^{T}{B, A}(U_{P})^{\dagger}{A, C}\hat{c}^{}{C} \right) \ & = \text{Tr}(H) - \sum_{C, D} \hat{c}^{\dagger}{D}H^{}{D, C}\hat{c}^{}_{C}, \end{split} \end{equation}
where $H$ is the first quantized Hamiltonian. For the Hamiltonian to respect PHS we require $\hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} = \hat{H}$ and therefore
\begin{equation} \begin{split} &\text{Tr}(H) = 0 \ U_{P}H^{*}&U_{P}^{\dagger} = -H. \end{split} \end{equation}
These conditions are complementary in the sense that the second condition implies the spectrum is symmetric about $E=0$ (try it!) which implies the first. 616