Let $S$ be a non-empty set and $\mathcal{A}$ a class of subsets of $S$ and define the following equivalence relation $\sim_{\mathcal{A}}$ on $S$;
$$s\sim_{\mathcal{A}}t\Longleftrightarrow 1_{A}(s)=1_{A}(t)\hspace{20pt}\forall A\in\mathcal{A}$$
Define $\mathcal{A}_{\tau}=\{[s]_{\sim{\mathcal{A}}}:s\in S\}$ to be the set of equivalence classes (the quotient set).
Question: Is it true that $\emptyset\in\mathcal{A}_{\tau}$??
Suppose $\emptyset\in\mathcal{A}_{\tau}$ and choose any $s,t\in\emptyset$. By definition of $\sim_{\mathcal{A}}$ this means $1_{A}(s)=1_{A}(t)$ for all $A\in\mathcal{A}$ vacuously, since of course $s,t$ do not exist. Thus no contradiction arises but this is not a proof that $\emptyset\not\in\mathcal{A}_{\tau}$ is true.
This seems related to the following question MSE question, but I wanted to make sure I was not being too lazy with applying this "vacuous" argument.
Showing that it would not be inconsistent if the empty set were in $\mathcal A_\tau$ does not prove that it is in $\mathcal A_\tau$, as you hint towards.
Look at the definition of $\mathcal A_\tau$. Is $\emptyset =[s]$ for some $s \in S$? Since we always have, at a minimum, $s\in [s]$, the answer is "No".
Notice that this doesn't depend on your particular equivalence. It's true for any equivalence relation. Even if $\emptyset$ was in your original set, its equivalence class, $[\emptyset]$, would, at a minimum, contain $\emptyset$, and even the set containing just the empty set is not the empty set.