Let $X$ be the intersection point of $AB, FH, CD$, and let $Y$ be the intersection point of $BC, EG, AD$. Let $L_1, L_2$ be the angle bisectors of $\angle AXO, \angle OXC$ respectively, and let $K_1, K_2$ be the angle bisectors of $\angle AYO, \angle OYC$ respectively.
Clearly, the centers of the inscribed circles of the four quadrilaterals (if they exist) must lie at the intersection points of $L_i, K_j$ $(i, j = 1, 2)$. Let $P_{ij}$ denote the intersection point of $L_i$ and $K_j$. We set
$$\alpha_1 = \frac{1}{2} \angle AXO, \quad \alpha_2 = \frac{1}{2} \angle OXC, \quad \beta_1 = \frac{1}{2} \angle AYO, \quad \beta_2 = \frac{1}{2} \angle OYC$$
The condition that the four quadrilaterals are tangential quadrilaterals is equivalent to the distance from $P_{ij}$ to one pair of opposite sides being equal to the distance to the other pair of opposite sides, which can be written as
$$XP_{ij} \sin \alpha_i = YP_{ij} \sin \beta_j$$
Now let $a_{ij} = XP_{ij} \sin \alpha_i, b_{ij} = YP_{ij} \sin \beta_j$, we want to prove that if $a_{ij} = b_{ij}$ holds for three of $(i, j) = (1, 1), (1, 2), (2, 1), (2, 2)$, then it also holds for the fourth.
Note that we have
$$\frac{XP_{11}}{XP_{12}} \cdot \frac{YP_{12}}{YP_{22}} \cdot \frac{XP_{22}}{XP_{21}} \cdot \frac{YP_{21}}{YP_{11}} = \frac{S_{XYP_{11}}}{S_{XYP_{12}}} \cdot \frac{S_{XYP_{12}}}{S_{XYP_{22}}} \cdot \frac{S_{XYP_{22}}}{S_{XYP_{21}}} \cdot \frac{S_{XYP_{21}}}{S_{XYP_{11}}} = 1$$

$$1 = \frac{a_{11}}{a_{12}} \cdot \frac{b_{12}}{b_{22}} \cdot \frac{a_{22}}{a_{21}} \cdot \frac{b_{21}}{b_{11}} = \frac{a_{11}}{b_{11}} \cdot \frac{a_{22}}{b_{22}} \cdot \frac{b_{12}}{a_{12}} \cdot \frac{b_{21}}{a_{21}}$$
Clearly, if three of the ratios on the right-hand side of the above equation are $1$, then the fourth is also $1$, which completes the proof!