By symmetry, it is useful to set $A(a, 0)$, $B(0, b)$. Denote $c = |OF| = \sqrt{a^2 - b^2}$ and the eccentricity $e = \frac{c}{a}$.
Let $|OP| = r$. By the properties of ellipse, we know that the range of $r$ is $[b, a]$.
Note that $O$ is the midpoint of $PQ$, and thus
$$\overrightarrow{FP} \cdot \overrightarrow{FQ} = (\overrightarrow{FO} + \overrightarrow{OP}) \cdot (\overrightarrow{FO} - \overrightarrow{OP}) = \overrightarrow{FO}^2 - \overrightarrow{OP}^2 = c^2 - r^2. \quad \textcircled{1}$$
(1) Suppose the focal point $F$ is $(c, 0)$, and then
$$\overrightarrow{FA} \cdot \overrightarrow{FB} = (a - c, 0) \cdot (-c, b) = c^2 - ac < 0.$$
Thus
$$\overrightarrow{FP} \cdot \overrightarrow{FQ} + \overrightarrow{FA} \cdot \overrightarrow{FB} < c^2 - r^2 < a^2 + b^2 = |AB|^2,$$
and this is not consistent with the conditions. Hence $F$ is $(-c, 0)$, which shows that $F$ lies on the extension of $AO$.
(2) From $F(-c, 0)$, we know that
$$\overrightarrow{FA} \cdot \overrightarrow{FB} = (a + c, 0) \cdot (c, b) = c^2 + ac.$$
Combining with (1) gives
$$\begin{aligned} \overrightarrow{FP} \cdot \overrightarrow{FQ} + \overrightarrow{FA} \cdot \overrightarrow{FB} &= (c^2 - r^2) + (c^2 + ac) \\ &= 2c^2 + ac - r^2, \end{aligned}$$
so $2c^2 + ac - r^2 = a^2 + b^2$, i.e.,
$$r^2 = 2c^2 + ac - a^2 - b^2 = 3c^2 + ac - 2a^2.$$
Note that $r \in [b, a]$, and thus $3c^2 + ac - 2a^2 = r^2 \in [a^2 - c^2, a^2]$.
The above equation is equivalent to $\frac{a^2 - c^2}{a^2} \le \frac{3c^2 + ac - 2a^2}{a^2} \le 1$, i.e.,
$1 - e^2 \le 3e^2 + e - 2 \le 1$.
Combining $e \in [0, 1]$, we obtain the range of the eccentricity of $\Gamma$, namely,
$$e \in \left[ \frac{3}{4}, \frac{-1 + \sqrt{37}}{6} \right].$$