Let $A = \{a_1, a_2, \dots, a_n\}$.
If there exist some $a_i > 1$, we can choose the largest one and, since $A$ is closed under multiplication, $a_i^2 \in A$, but $a_i^2>a_i$, thus contradicting that $a_i$ is the maximum number in $A$. Similarly, $a_i \ge -1$, for every $i \in \{1,2, \dots, n\}$.
Then, obviously, $a_1a_2\dots a_n \in A$, so there is a $j \in \{1, 2, ..., n\}$ such that $a_j = a_1a_2 \dots a_n$. If $a_j \neq 0$, $a_1a_2 \dots a_{j-1}a_{j+1} \dots a_n = 1$, so $a_1a_2 \dots a_{i-1}a_{i+1} \dots a_{j-1}a_{j+1} \dots a_n = \frac{1}{a_i} \in A$, for every $i \in \{1, 2, \dots j - 1, j + 1 \dots, n\}$. If $a_j = 0$, we can a similar judgement for $A \setminus \{a_j\}$.
If we suppose $a_i \in (-1, 1) \setminus \{0\}$, $i \in \{1, 2, \dots j - 1, j + 1, \dots, n\}$, then $\frac{1}{a_i} \in (-\infty, 1) \cup (1, \infty)$, which is absurd, since we proved that every elements of A is in $[-1, 1]$. So, $a_i \in \{-1, 0, 1\}$, for every $i \in \{1, 2, \dots, j - 1, j + 1 \dots, n\}$. So, $A = \{-1, 0, 1, a_j\}$. It is easy to see that $a_j \in \{-1, 0, 1\}$ too.
After checking every subset of $\{-1, 0, 1\}$, we arrive at the conclusion that: $$A \in \{\{0\}, \{1\}, \{0, 1\}, \{-1, 1\}, \{-1, 0, 1\}\}$$
I hope my solution is correct, as I am not too sure about the part where I proved $a_i \in \{-1, 0, 1\}$, for every $i \in \{1, 2, \dots, j - 1, j + 1, \dots n\}$