Solution:
Suppose $0 \in A$. Then $a^{2} = a^{2} + 0 \times b$ is rational and $a b = 0^{2} + a b$ is also rational for all $a, b$ in $A$, $a \neq 0$, $b \neq 0$, $a \neq b$. Hence $a = a_{1} \sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \neq 0$, we have
$$b \sqrt{M} = \frac{a b}{a_{1}}$$
which is a rational number.
Hence we may assume $0$ is not in $A$. If there is a number $a$ in $A$ such that $-a$ is also in $A$, then again we can get the conclusion as follows. Consider two other elements $c, d$ in $A$. Then $c^{2} + d a$ is rational and $c^{2} - d a$ is also rational. It follows that $c^{2}$ is rational and $d a$ is rational. Similarly, $d^{2}$ and $c a$ are also rationals. Thus $d / c = (d a) / (c a)$ is rational. Note that we can vary $d$ over $A$ with $d \neq c$ and $d \neq a$. Again $c^{2}$ is rational implies that $c = c_{1} \sqrt{M}$ for some rational $c_{1}$ and natural number $M$. We observe that $c \sqrt{M} = c_{1} M$ is rational, and
$$a \sqrt{M} = \frac{c a}{c_{1}}$$
so that $a \sqrt{M}$ is a rational number. Similarly is the case with $-a \sqrt{M}$. For any other element $d$,
$$b \sqrt{M} = M c_{1} \frac{d}{c}$$
is a rational number.
Thus we may now assume that $0$ is not in $A$ and $a + b \neq 0$ for any $a, b$ in $A$. Let $a, b, c, d$ be four distinct elements of $A$. We may assume $|a| > |b|$. Then $d^{2} + a b$ and $d^{2} + b c$ are rational numbers and so is their difference $a b - b c$. Writing $a^{2} + a b = a^{2} + b c + (a b - b c)$, and using the facts $a^{2} + b c$, $a b - b c$ are rationals, we conclude that $a^{2} + a b$ is also a rational number. Similarly, $b^{2} + a b$ is also a rational number.
Consider
$$q = \frac{a}{b} = \frac{a^{2} + a b}{b^{2} + a b}$$
Note that $a^{2} + a b > 0$. Thus $q$ is a rational number and $a = b q$. This gives $a^{2} + a b = b^{2}(q^{2} + q)$. Let us take $b^{2}(q^{2} + q) = l$. Then
$$|b| = \sqrt{\frac{l}{q^{2} + q}} = \sqrt{\frac{x}{y}}$$
where $x$ and $y$ are natural numbers. Take $M = x y$. Then $|b| \sqrt{M} = x$ is a rational number. Finally, for any $c$ in $A$, we have
$$c \sqrt{M} = b \sqrt{M} \frac{c}{b}$$
is also a rational number.