From $\frac{1}{a} + \frac{1}{b} \le 2\sqrt{2}$, we have $a + b \le 2\sqrt{2}ab$. On the other hand,
$$(a + b)^2 = 4ab + (a - b)^2 = 4ab + 4(ab)^3 \\ \ge 4 \cdot 2\sqrt{ab} \cdot (ab)^3 = 8(ab)^2,$$
and that means
$$a + b \ge 2\sqrt{2}ab. \qquad \textcircled{1}$$
Therefore,
$$a + b = 2\sqrt{2}ab. \qquad \textcircled{2}$$
The equality in 1 holds only when $ab = 1$. Associating it with 2, we find
$$\begin{cases} a = \sqrt{2} - 1, \\ b = \sqrt{2} + 1, \end{cases} \quad \text{and} \quad \begin{cases} a = \sqrt{2} + 1, \\ b = \sqrt{2} - 1. \end{cases}$$
So the answer is $\log_a b = -1$.