Prove or disprove that $PQ = P + Q - I$ if $P$ and $Q$ are disjoint permutation matrices whose cycle lengths sum to $n.$

Here's a way to transform your great illustration into a proof:

Let $e_1,\dots,e_n$ be the standard basis of $\Bbb R^n$, then $P$ and $Q$ act on these exactly as the corresponding permutations.

So, by the condition, each $e_i$ is either moved by $P$ and fixed by $Q$, or is moved by $Q$ and fixed by $P$, moreover then $Qe_i$ is still in the cycle of $Q$, thus it's also fixed by $P$.

In the first case we have $Qe_i=e_i$, $$PQe_i=Pe_i=Pe_i+e_i-e_i=(P+Q-I)e_i\,.$$ While in the second case we have $Pe_i=e_i$ and $PQe_i=Qe_i$, $$PQe_i=Qe_i=e_i+Qe_i-e_i=(P+Q-I)e_i\,.$$ Since this holds for each element of a basis, it holds for all vectors, and hence $PQ=P+Q-I$.

(Note that this proof also allows $P$ and $Q$ to have more cycles, and the condition that we really need here is that the sets of fixed points of $P$ and $Q$ are disjoint.)

An alternative approach: note that $$ PQ = P + Q - I \iff PQ - P - Q + I = 0 \iff (P - I)(Q - I) = 0. $$ From there, note that if $P$ corresponds to a permutation of the elements of $M \subset \{1,\dots,n\}$ (i.e. all elements not in $M$ are fixed), then we will have $(P - I)e_i = 0$ for all $i \notin M$.