First solution. Transform the original expression:
$$\begin{aligned} 0 &= a^3 b + ab^3 + 2a^2 b^2 + 2a + 2b + 1 = ab(a + b)^2 + 2(a + b) + 1 \\ &= (ab - 1)(a+b)^2 + (a+b)^2 + 2(a+b) + 1 = (ab - 1)(a+b)^2 + (a+b+1)^2. \end{aligned}$$
Notice that $a + b \neq 0$, because otherwise $1 = 0$.
Therefore, $1 - ab = \left(\frac{a+b+1}{a+b}\right)^2$; since $a$ and $b$ are rational numbers, $\frac{a+b+1}{a+b}$ is also a rational number, and the statement is proved.

Second solution. Multiply the given equation by $ab$ and transform:
$$a^2b^2(a+b)^2 + 2ab(a+b) + ab = 0 \Leftrightarrow a^2b^2(a+b)^2 + 2ab(a+b) + 1 = 1 - ab \Leftrightarrow 1 - ab = (ab(a+b) + 1)^2.$$
From the last equality, the statement follows.

Third solution. If $ab = 0$, then the statement is true:
$$1 - ab = 1^2.$$
If $ab \neq 0$, then the quadratic equation $abx^2 + 2x + 1 = 0$ has a rational root ($a + b$) and rational coefficients. By the quadratic formula, its discriminant is a square of a rational number. But $D = 4(1 - ab)$, and the statement is proved.