Prove or disprove that $PQ = P + Q - I$ if $P$ and $Q$ are disjoint permutation matrices whose cycle lengths sum to $n.$ 
Prove or disprove that if the matrices $P$ and $Q$ represent disjoint permutation cycles in $S_{n}$ with sum of cycle lengths equal to $n,$ then $PQ = P+Q-I$.

MY TRY:
Let's start by an example. Let $P$ and $Q$ be the matrices corresponding to the respective permutations $p = (1 \, 2)$ and $q = (3 \, 4 \, 5)$ in cycle notation. We have that $$
P =
\begin{pmatrix}
0 & 1 & 0 &  0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 &  0 & 0\\ 
0 & 0 & 0 &  1 & 0\\
0 & 0 & 0 &  0 & 1
\end{pmatrix} \text{ and } Q = \begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0\\ 
0 & 0 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.
$$ It seems obvious that the matrix $PQ$ representing the permutation $pq = (1 \, 2)(3 \, 4 \, 5)$ will be $P+Q-I,$ as the "untouched" $1$s in the matrices are simply canceled by $I$ and the touched $1$s create the derangements. But isn't there some clear method to prove it?
I am new to group theory. Please ask for clarifications in case of any discrepancies. Any hint will be a great help!

Not an answer but a generalisation of the above:-

Let $P_{1}, P_{2},\cdots P_{n}$ ($n\gt 2$) represent disjoint permutation cycles, we'll have $$\prod_{i=1}^{n}P_{i} = \sum_{i=1}^{n}P_{i}-(n-1)I$$

Proof :- We'll prove it by induction. It holds for $n=2$, as proved in the problem. Let it holds for $n = k$ i.e. $$\prod_{i=1}^{k}P_{i} = \sum_{i=1}^{k}P_{i}-(k-1)I$$ we'll prove it also holds for $n=k+1$. Let $P_{k+1}$ represent permutation cycle disjoint to $P_{1},P_{2}\cdots P_{k}$ then  $$\biggl(\prod_{i=1}^{k}P_{i}\biggr)P_{k+1} = P_{1}P_{k+1}+P_{2}P_{k+1}+\cdots P_{k}P_{k+1}-(k-1)P_{k+1}.$$ We also have $$P_{1}P_{k+1} = P_{1}+P_{k+1}-I $$ $$P_{2}P_{k+1} = P_{2}+P_{k+1}-I$$ $$ \vdots$$ $$P_{k}P_{k+1} = P_{k}+P_{k+1}-I$$ adding these equations we get $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1} = \sum_{i=1}^{n}P_{i}+kP_{k+1}-kI$$ $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1}-(k-1)P_{k+1} = \sum_{i=1}^{n}P_{i}+kP_{k+1}-kI-(k-1)P_{k+1}$$ $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1}-(k-1)P_{k+1} = \sum_{i=1}^{k+1}P_{i}-kI$$  which gives $$\prod_{i=1}^{k+1}P_{i}=\sum_{i=1}^{k+1}P_{i}-kI$$
