The answer is no. Consider the space $X=\{0,1,2\}$ with the topology $\tau=\{\emptyset,\{0\},\{1,2\},X\}$. There are precisely two injective neighborhood selectors, both of which are almost surjective in your sense. Namely, we must map $0\mapsto\{0\}$ and then $1$ and $2$ get mapped to $\{1,2\}$ and $X$, in either way.

So this space is critical in your sense, but the topology is not of the form $T_\alpha$ for any ordinal $\alpha$. For example, it is not $T_0$, since we cannot separate $1$ and $2$.

The answer is no. Consider the space $X=\{0,1,2\}$ with the topology $\tau=\{\emptyset,\{0\},\{1,2\},X\}$. There are precisely two injective neighborhood selectors, both of which are almost surjective in your sense. Namely, we must map $0\mapsto\{0\}$ and then $1$ and $2$ get mapped to $\{1,2\}$ and $X$, in either way.

So this space is critical in your sense, but the topology is not of your form $\tau_\alpha$ for any ordinal $\alpha$. For example, it is not $T_0$, since we cannot separate $1$ and $2$.