The given condition is equivalent to
$$(ax^2 y^2 + b) + (a(x + y)^2 + b) \ge (ax^2 + b)(ay^2 + b). \quad ①$$
In ①, let $y = 0$. We have $b + (ax^2 + b) \ge (ax^2 + b) \cdot b$, or
$$(1 - b)ax^2 + b(2 - b) \ge 0.$$
As $a > 0$ and $ax^2$ can be sufficiently large, then $1 - b \ge 0$, i.e., $0 < b \le 1$.
In ①, let $y = -x$. We have $(ax^4 + b) + b \ge (ax^2 + b)^2$, or
$$(a - a^2)x^4 - 2abx^2 + (2b - b^2) \ge 0. \quad ②$$
Denote the left-hand side of ② as $g(x)$. It is obvious that $a - a^2 \ne 0$ (otherwise, from $a > 0$ we know $a = 1$. Then $g(x) = -2bx^2 + (2b - b^2)$ with $b > 0$, which means $g(x)$ can be negative. A contradiction). Then
$$\begin{align*} g(x) &= (a - a^2) \left( x^2 - \frac{ab}{a - a^2} \right)^2 - \frac{(ab)^2}{a - a^2} + (2b - b^2) \\ &= (a - a^2) \left( x^2 - \frac{b}{1-a} \right)^2 + \frac{b}{1-a} (2 - 2a - b) \\ &\ge 0 \end{align*}$$
holds for any real number $x$. So we have $a - a^2 > 0$, i.e., $0 < a < 1$.
Furthermore, from $\frac{b}{1-a} > 0$ and
$$g\left(\sqrt{\frac{b}{1-a}}\right) = \frac{b}{1-a}(2 - 2a - b) \ge 0,$$
we have $2a + b \le 2$.
So far, we get the necessary condition that $a, b$ must satisfy as follows:
$$0 < b \le 1, \quad 0 < a < 1, \quad 2a + b \le 2. \qquad \textcircled{3}$$
We are going to prove that for any pair $(a, b)$ satisfying ③ and any real numbers $x, y$, ① holds, or equivalently,
$$h(x, y) = (a - a^2)x^2 y^2 + a(1 - b)(x^2 + y^2) + 2axy + (2b - b^2) \ge 0.$$
As a matter of fact, when ③ holds, we then have
$$a(1-b) \ge 0, \quad a - a^2 > 0 \quad \text{and} \quad \frac{b}{1-a}(2 - 2a - b) \ge 0.$$
Combining it with $x^2 + y^2 \ge -2xy$, we get
$$\begin{align*} h(x, y) &\ge (a - a^2)x^2 y^2 + a(1 - b)(-2xy) + 2axy + (2b - b^2) \\ &= (a - a^2)x^2 y^2 + 2abxy + (2b - b^2) \\ &= (a - a^2) \left( xy + \frac{b}{1-a} \right)^2 + \frac{b}{1-a} (2 - 2a - b) \ge 0. \end{align*}$$
Therefore, the set of all the pairs $(a, b)$ meeting the given condition is
$$\{(a, b) \mid 0 < b \le 1, 0 < a < 1, 2a + b \le 2\}.$$