Let $a_1 > a_2 > a_3 > a_4$ be the four largest elements in $A$. Let $a = a_1a_2 + a_3a_4$, $b = a_1a_3 + a_2a_4$ and $c = a_1a_4 + a_2a_3$. Then $a > b > c$ by the rearrangement inequality. We also know that $a, b, c \in A$.

If there is a fifth term $a_5$, then as before, the 4 terms $a_2, a_3, a_4, a_5$ generate 3 numbers $a' > b' > c'$ and we have $c = a_1a_4 + a_2a_3 > a' = a_2a_3 + a_4a_5$. Thus we have 6 distinct numbers. In general, if we have $n + 4$ terms, $n \ge 1$, then every 4 consecutive terms generate 3 distinct elements, giving a total of $\ge 3(n+1) > n + 4 = |A|$. Thus $A = \{a_1, a_2, a_3, a_4\}$.

Suppose $a_4 > 0$. If $c = a_4$, then
$$(1 - a_1)a_4 = a_2a_3 \Rightarrow 1 - a_1 > a_2 \Rightarrow a_2 < \frac{1}{2} \Rightarrow a_4 < \frac{1}{2} \Rightarrow a < \frac{a_1+a_3}{2} < a_1.$$
Hence $a = a_2$. Therefore
$$1 = a_1 + \frac{a_3a_4}{a_2} = a_1 + \frac{a_2a_3}{a_4} \Rightarrow a_2 = a_4, \text{ contradiction.}$$
If $c > a_4$, then $a_1 = a = a_1a_2 + a_3a_4$, $a_2 = b = a_1a_3 + a_2a_4$, $a_3 = c = a_1a_4 + a_2a_3$. Therefore
$$1 = a_2 + \frac{a_3a_4}{a_1} = a_2 + \frac{a_1a_4}{a_3} \Rightarrow a_4 = 0, \text{ contradiction.}$$
Hence $a_4 = 0$ and since $c > 0$, $c = a_3$, $b = a_2$ and $a = a_1$. Thus
$$a_3 = a_4a_1 + a_2a_3 = a_2a_3 \Rightarrow a_2 = 1; \quad a_2 = a_1a_3 + a_2a_4 \Rightarrow a_1a_3 = 1.$$
It is easy to check that $A = \{x, 1, \frac{1}{x}, 0\}$, satisfies the conditions given.