Why is a minimum field configuration called a vacuum state in SSB?

To call a classical extremum of the field a "vacuum state" is a common abuse of terminology. It is not the extremum itself that is a vacuum state, but to every classical extremum there is (to first order) an associated vacuum.

Since the LSZ formalism and the Feynman rules need vanishing vacuum expectation value (VEV) to work, we need to rewrite fields with non-zero VEV in terms of perturbations about their VEV, i.e. $\tilde{\phi} := \phi - \langle 0 \vert \phi \vert 0\rangle$ are our dynamical fields. To first order, the VEV is well-approximated by the classical minimum, cf. this question and its answers and this answer by Prahar. At weak couplings, perturbative renormalization can ensure that the VEV of the perturbed field is zero order by order in perturbation theory (this should be shown in e.g. Coleman's Aspects of Symmetry).

Now, to each different VEV there must belong a different vacuum state - we can't have the same state and the same field giving two different VEVs, and there is nothing else in that expression the VEV could depend on. (Note that whether you say the vacuum state or the representation of the field as an operator is different doesn't matter, you get "different" Hilbert spaces (perturbatively) built on the vacuum state in both cases.) We also should concede that the existence of such vacua associated to each VEV is a postulate of QFT - since we can rarely explicitly construct the space of states of a QFT anyway, it seems infeasible to give any proof of existence of these vacua.