I have the following queries about the lemma on saturated non-factorizable graphs. My attempt at constructing a saturated non-factorizable graph also seems to contradict the lemma.
(The proof is taken from the book A walk Through Combinatorics by Miklos Bona.)
Just look at the definition of symmetric difference!
The cycle $C_1$ is an alternating (and therefore even) cycle and $F_3= (F_1 \cup C_1) \setminus (F_1 \cap C_1)$. So you actually interchange the edges of $C_1$ (so the edges of $C_1$ which are in $F_1$ are not in $F_3$ and vice versa). Therefore $F_1$ and $F_3$ have the same number of edges.
We don't need that the whole of $P\cup ab$ is not in $F_2$, why do you want it? We have $F_4=(F_2 \cup (P\cup ab)) \setminus (F_2 \cap (P\cup ab))$, so we again interchange the edges of the (even) cycle $P\cup ab$ (so the edges of $P\cup ab$ which are in $F_2$ are not in $F_4$ and vice versa). That's why $F_2$ and $F_4$ have the same number of edges.
Why does your example contradict the Lemma? You have $S=\{A, B\}$ and the components of $G-S$ are the four components $C_1=C$, $C_2=D$, $C_3=E$ and $C_4=F$ (which are all complete graphs). So everything is okay.