The short answer is that it depends on which of the symmetries you enforce. More precisely, the simple SSH model has many symmetries, and it is a priori not clear which of these symmetries you consider as 'accidental' and which you consider as 'enforced'. This is a matter of choice. Depending on this choice, the model lands up in different possible symmetry classes (possible choices are A, AIII, AI, BDI and D, as I will explain; the respective invariants are $0$, $\mathbb Z$, $0$, $\mathbb Z$, $\mathbb Z_2$).
Let me give some more detail. Consider the SSH model
$$ H_\textrm{SSH} = - \sum_n \left( t_{AB} \; c^\dagger_{n,A} c_{n,B} + t_{BA} \; c_{n,B}^\dagger c_{n+1,A} + h.c. \right). $$
In the limit $t_{BA} = 1$ and $t_{AB} = 0$, we obtain the fixed point limit where the coupling is purely between unit cells, giving rise to a decoupled zero-energy fermion at each end of the chain.
It is conventional to define a single-particle Hamiltonian $\mathcal H_k$ through
$$ H = \sum_k \left( c^\dagger_{k,A}, c^\dagger_{k,B} \right) \mathcal H_k \left( \begin{array}{c} c_{k,A} \\ c_{k,B} \end{array} \right).$$
Hence for $H = H_\textrm{SSH}$, we have that $\mathcal H_{\textrm{SSH},k} = -\left[t_{AB} + t_{BA} \cos(k) \right] \sigma_x - t_{BA} \sin(k) \sigma_y$.
This model has a lot of symmetries. Let me go through them, focusing on how they act on the single-particle Hamiltonian (*):
- A commuting anti-unitary 'time-reversal' symmetry $\mathcal T$ defined by $\mathcal T \mathcal H_k \mathcal T := \mathcal H_{-k}^*$ and $\mathcal T^2 = +1$. We see that $[\mathcal T, \mathcal H_{\textrm{SSH},k}] = 0$.
- An anti-commuting unitary 'sublattice' symmetry $\mathcal S$ defined by $\mathcal S \mathcal H_k \mathcal S := \sigma_z \mathcal H_k \sigma_z$ and $\mathcal S^2 = +1$. We see that $\{ \mathcal S, \mathcal H_{\textrm{SSH},k} \} = 0$.
- An anti-commuting anti-unitary `particle-hole' symmetry $\mathcal C$. We can simply define $\mathcal C = \mathcal S \mathcal T$. We have that $\mathcal C^2 = +1$ and $\{ \mathcal C, \mathcal H_{\textrm{SSH},k} \} = 0$.
Hence we see that the SSH model has all the three symmetries $\mathcal T,\mathcal C,\mathcal S$ that enter the periodic table of topological insulators/superconductors! We thus get to choose which class we put it in. You might think 'if it has all symmetries, then we must put it in the class BDI, which has all three symmetries'. That is not quite true: the class is not defined by 'which symmetries does our model have?' but rather 'what kind of arbitrary symmetric terms do we allow to add to our model?'. Let me give some more detail.
"The SSH model is in the class AIII": if we say this, we mean that we allow any perturbations that respect $\mathcal S$, but they need not obey $\mathcal T$ and $\mathcal C$. The table tells us there infinitely many distinct gapped phases, labeled by an integer $\mathbb Z$. This is easy to understand: the $\mathcal S$-symmetry above tells us that $\mathcal H_k$ has to anticommute with $\sigma_z$, hence $\mathcal H_k = h_x(k) \sigma_x + h_y(k) \sigma_y$. Since our model is gapped, we have a well-defined map $$S^1 \to \mathbb R^2 - \{0\}: k \to (h_x(k),h_y(k)).$$
This is an embedding of the circle into the punctured plane, which has a well-defined winding number around the origin. One can prove that the winding number is equivalent to
$$ \nu = \frac{1}{\pi} \int_{-\pi}^\pi \mathrm dk\ \; \langle \psi_k| \sigma_z i \partial_k |\psi_k \rangle. $$
It is straight-forward to derive that for the SSH model, we have $\nu = 0$ if $t_{AB} > t_{BA}$ (trivial phase) and $\nu = 1$ for $t_{AB} < t_{BA}$ (topological phase). The classification tells us that no matter what $\mathcal S$-symmetric term we add, we cannot adiabatically connect these two gapped phases.
"The SSH model is in the class BDI": this means we enforce all three symmetries. Since we saw that $\mathcal S$ by itself was already enough to protect $\mathbb Z$ distinct phases, it is trivial to observe that with extra symmetries, our classification does not get smaller.
"The SSH model is in the class D": this means we allow all perturbations that respect $\mathcal C$, but they can break $\mathcal T$ and/or $\mathcal S$. One can show that one can now connect a model that has $\nu = 2$ to one that has $\nu =0$. In the class AIII we could not do this. More generally, it turns out only $\nu \mod 2$ is a well-defined invariant (i.e. this number cannot change without a phase transition). Since the SSH model had $\nu = 1$ in the topological phase, we see that it is still a non-trivial phase in the class D. Equivalently, this $\mathbb Z_2$ invariant can be measured by
$$ \gamma = \frac{1}{\pi} \int_{-\pi}^\pi \mathrm dk\ \; \langle \psi_k| i \partial_k |\psi_k \rangle. $$
Indeed, one can show that $\gamma \equiv \nu \mod 2$.
"The SSH model is in the class A or AI": now we allow all possible term (class A) or all $\mathcal T$-preserving terms (class AI). The classification tells us that in either case, we can smoothly connect all gapped models. Indeed, nothing prevents us from adding an on-site potential to the SSH model, which can be used to smoothly connect the limit $t_{AB}=0,t_{BA}=1$ to the limit $t_{AB}=1,t_{BA}=0$. Hence, we are allowed to say that the SSH model is in one of these two classes, but if we do so, its edge modes are no longer topologically protected.
(*) Note that it is actually more natural (but, alas, less conventional) to consider how the symmetries act on Fock space, i.e. how they act on the 'actual' hamiltonian $H$. Then the three symmetries $T$, $C$ and $S$ are all commuting, as one would desire of a symmetry! It is only one when considers their effective action on the single-particle Hamiltonian $\mathcal H_k$ that some become anti-commuting, in an affront to our physical intuition. More precisely, $T$ is defined to complex-conjugation in the physical occupation basis. However, $C$ is defined to a unitary (commuting) symmetry via $c_{n,A} \leftrightarrow c^\dagger_{n,A}$ and $c_{n,B} \leftrightarrow -c^\dagger_{n,B}$. Note that this naturally explains its name as a 'particle-hole' transformation. The reason it seems to act as an anti-commuting anti-unitary symmetry $\mathcal C$ on the single-particle Hamiltonian has to do with it interchanging daggers. The latter can be rewritten as a transpose, up to a sign. Using hermiticity, the transpose can be replaced by complex-conjugation.
