$a_{\min} = \frac{1+\sqrt{5}}{2}$.

Write $\varphi = \frac{1+\sqrt{5}}{2}$. We may assume that each edge has length $\sqrt{2}$. For any point $P$ inside the square $ABCD$, let $S_1, S_2, S_3, S_4$ denote the area of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$ respectively; we can also assume that $S_1 \ge S_2 \ge S_3 \ge S_4$.

Let $\lambda = \frac{S_1}{S_2}$, $\mu = \frac{S_2}{S_4}$. If $\lambda, \mu > \varphi$, as
$$S_1 + S_3 = S_2 + S_4 = 1,$$
we have $\frac{S_2}{1-S_2} = \mu$, $S_2 = \frac{\mu}{1+\mu}$. So

$$S_1 = \lambda S_2 = \frac{\lambda \mu}{1+\mu} = \frac{\lambda}{1+\frac{1}{\mu}} > \frac{\varphi}{1+\frac{1}{\varphi}} = \frac{\varphi^2}{1+\varphi} = 1,$$
and we reach a contradiction. Hence, $\min\{\lambda, \mu\} \le \varphi$, which implies that $a_{\min} \le \varphi$.

On the other hand, for any $a \in (1, \varphi)$, we take any $t \in (a, \frac{1+\sqrt{5}}{2})$ such that $b = \frac{t^2}{1+t} > \frac{8}{9}$. Inside the square $ABCD$ we can choose a point $P$ so that $S_1 = b$, $S_2 = \frac{b}{t}$, $S_3 = \frac{b}{t^2}$, $S_4 = 1-b$. Then we have
$$\frac{S_1}{S_2} = \frac{S_2}{S_3} = t \in (a, \frac{1+\sqrt{5}}{2}),$$
$$\frac{S_3}{S_4} = \frac{b}{t^2(1-b)} > \frac{b}{4(1-b)} > 2 > a.$$
Thus, for any $i, j \in \{1, 2, 3, 4\}$, we have $\frac{S_i}{S_j} \notin [a^{-1}, a]$.

Hence $a_{\min} = \varphi$.