We clear denominators to get
$$\begin{align*} & a(a+1) + b(b+1) \le (a+1)(b+1), \\ \Leftrightarrow \quad & a^2 + a + b^2 + b \le ab + a + b + 1, \\ \Leftrightarrow \quad & a^2 - a + b^2 - b \le ab - a - b + 1, \\ \Leftrightarrow \quad & a(a-1) + b(b-1) \le (a-1)(b-1), \\ \Leftrightarrow \quad & (1-a)(1-b) + a(1-a) + b(1-b) \ge 0. \end{align*}$$
The three terms on the left-hand side of the last inequality are clearly all positive or zero for $0 \le a, b \le 1$.
For equality to hold, all three terms have to be zero, that is, $a = 1$ or $b = 1$ and $a, b \in \{0, 1\}$.
This gives the three pairs $(a, b) = (1, 0)$, $(a, b) = (0, 1)$ and $(a, b) = (1, 1)$.