First of all, the SSH model does not have particle-hole symmetry. Particle-hole symmetry is an exact symmetry (at mean field level) reserved for superconductors and is an anti-unitary symmetry. A symmetric spectrum does not mean particle-hole symmetry.
SSH model and end states
Let me first explain a simple way to understand the topological end states. The SSH model is a 1D chain of dimers on which live spinless electrons that can hop from site to site:
\begin{equation}
H = \sum_n \underbrace{t_1 c_{An}^\dagger c_{Bn}}_{\textrm{intracell}} + \underbrace{t_2 c_{An+1}^\dagger c_{Bn}}_{\textrm{intercell}} + \,h.c,
\end{equation}
with bulk energy dispersion $E(k) = \left(t_1 + t_2 \cos k\right)^2 + \left( t_2 \sin k \right)^2$.
Consider the strong-coupling limits:
- $t_1 \gg t_2$: This a molecular limit where all atoms in the chain form dimers. The spectrum of this chain is just $E = \pm t_1$. Clearly trivial. No end states.
- $t_2 \gg t_1$: Now the atoms at the end of the chain are unpaired. Since the model has zero on-site energy, an electron on these ends has zero energy. The bulk spectrum is $E = \pm t_2$ with two localized zero modes, one at each end.
Symmetries and protection
Note that a spectrum that is symmetric around zero energy is not necessary for the end states. The end states will just not lie at zero energy. In fact, the symmetric spectrum is artificial, since second-nearest hoppings introduce an asymmetry.
However, what is a real symmetry of the chain, is inversion symmetry about the centre of a bond and spinless time-reversal symmetry. Moreover, the "topology" of the SSH system is protected by these symmetries. In fact, it gives rise to a $\mathbb Z_2$ classification.
No change in the Hamiltonian that respects inversion symmetry $H(-k) = \sigma_x H(k) \sigma_x$ and spinless time-reversal symmetry $H(-k) = H(k)^*$ and does not close
the energy gap can destroy the end modes in the topological regime ($t_2 \gg t_1$). Note that this forbids $\sigma_z$.
Dirac Hamiltonian
Recall the strong-coupling regime. In order to go from the trivial regime to the topological regime, you need to go through the point $t_1 = -t_2$ at which point the energy gap closes at $k=0$. So in order to change topology, the system has to become gapless. Near $k=0$, the Bloch Hamiltonian of the SSH model is given by the Dirac Hamiltonian:
\begin{equation}
h = m \sigma_x + t_2 k \rightarrow m(x) \sigma_x - it_2 \partial_x \sigma_y,
\end{equation}
where $m(x) = t_1(x) + t_2$. At the closing of the gap $m=0$. Now consider a domain wall between two different phases (see figure below). At a domain wall $m(x)$ changes sign. In this case, there exists a normalizable zero mode localized at the domain wall:
\begin{equation}
h \psi = 0 \Rightarrow \psi(x) = \exp \left( -\int_a^x dx' \frac{m(x')}{t_2} \right) \begin{pmatrix} 1 \\ 0 \end{pmatrix},
\end{equation}
where the lower limit $a$ fixes the normalization. This is the Jackiw-Rebbi model. You see that the condition for a zero mode requires a change of the sign of the mass $m(x)$. This mechanism, called band inversion, is generic for all symmetry-protected topological states such as topological insulators, topological superconductors, crystalline topological insulators, $\ldots$.
In essence, the symmetries were necessary to restrict $h$ to only two Pauli matrices. In order to have a generic gap closing in 1D you can only have two Pauli matrices. Any old symmetry that forbids one of the three Pauli matrices will do. If you allow for three Pauli matrices, you allow for two independent parameters apart from the momentum $k$ and any perturbation can prevent the gap from closing when going between the phases (which means no zero mode). To see this, note that for two parameters you can make a detour away from the gap closing point. This is basically the Neumann-Wigner theorem.
