No, in general permutation matrices do not commute.

It seems like you are using the Doolittle algorithm, so I am going to assume that indeed you are.

Let $U_i$ be the $i$:th step in your LU decomposition of $A$, so
$$\begin{align}
U_0 &= A \\
U_1 &= L_1P_1U_0 \\
\vdots \\
U_n &= L_nP_nU_{n-1}
\end{align}$$

This corresponds well to how one would do LU-decomposition by hand; get the largest element as the leading element on the row you are at (i.e. multiply with $P_k$), then eliminate that column on the following rows (i.e. multply with $L_k$).

As you remark, the $L_i$ will be atomic lower triangular matrices, the non-zero elements all being in column $i$. The inverse of $L_i$ can be constructed by negating all off-diagonal elements, see Wikipedia.

The permutation matrix $P_j$ will be a permutation matrix switching row $j$ with a row $k \geq j$, if multiplied on the left of the matrix you want to transform. The inverse to $P_j$ is $P_j$ itself (since $P_j$ switches row $j$ with row $k$, you can undo this by doing the same thing).

Consider the product $P_jL_i$ for $i < j$. $P_j$ will switch two rows of $L_i$, row $j$ and $k \geq j > i$. We switch elements in $L_i$ as follows:
$$\begin{align}
L_{j,i} &\leftrightarrow L_{k,i} \\
L_{j,j} &\leftrightarrow L_{k,j}
\end{align}$$
Let $L_k'$ be the matrix produced by switching just the off-diagonal elements (the first switch above). Note that this is still an atomic lower triangular matrix. We can then produce $P_jL_i$ by just switching **column** $j$ with column $k$ in $L_k'$, which is multiplying by $P_j$ on the **right**. Here it is important that $i < j, k$, so column $i$ in $L_i'$ is not changed! In other words:
$$P_j L_i = L_i' P_j$$

Thus, you can from your equation
$$L_nP_nL_{n-1}...P_3L_2P_2L_1P_1A = U$$
produce
$$L_n^S L_{n-1}^S \cdots L_1^S P_n P_{n-1} \cdots P_1A = U$$
which can be transformed to (note that $L_i^S$ is still atomic lower triangular):
$$PA = LU$$
taking
$$P = P_nP_{n-1} \cdots P_1$$
and
$$L = (L_n^S L_{n-1}^S \cdots L_1^S)^{-1}.$$

Here, $L_i^S$ is the matrix made by taking $L_i$ and applying all $P_j$ (on the left) for $j > i$ on the **off-diagonal elements**.

You do this by doing the following, starting with
$$L_nP_nL_{n-1}...P_3L_2P_2L_1P_1A = U$$
move the $P_2$ matrix to the right using $P_2L_1 = L_1' P_2$, producing:
$$L_nP_nL_{n-1}...P_3L_2L_1'P_2P_1A = U$$
then do the same for the matrix $P_3$, but this matrix you have to move to the right twice, using $P_3L_2 = L_2'P_3$ and $P_3L_1' = L_1''P_3$:
$$L_nP_nL_{n-1}...L_2'L_1''P_3P_2P_1A = U$$
and so on for every $P_j$.

As an example of $P_j L_i = L_i' P_j$ consider
$$P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$
$$L = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{pmatrix}$$
$$L' = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}$$

$$PL = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix}$$
$$L'P = \begin{pmatrix}1 & 0 & 0 \\ 3 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix}$$

So, to summarize: These special matrices *almost* commute, only small changes are needed to swap the matrices. However, all important properties of the matrices are preserved are when swapping them.