Is there an intuitive proof of the identity $ \sum_{L \subset S} \prod_{x \in L} (x-1) = \prod_{x \in S} x$ from general principles?

Let $S'$ be $\{x : x-1 \in S\}$, in other words we shift $S$ by one.

The equality $$ \sum_{L \subset S} \prod_{x \in L} (x-1) = \prod_{x \in S} x$$

reduces to

$$ \sum_{L \subset S'} \prod_{x \in L} x = \prod_{x \in S'} (x+1)$$

but this is obviously distributivity: LHS is the expanded RHS.

If $S=\emptyset$, then $L=\emptyset$ is the only subset and by convention the empty product $\prod_{x\in L} (x-1)$ equals $1$. That shows $$\tag1 \sum_{L\subseteq S}\prod_{x\in L}(x-1)=\prod_{x\in S}x$$ holds at least if $S=\emptyset$. For an induction step, let $a\in S$ be any element and let $S'=S\setminus\{a\}$. By assumption, $(1)$ holds with $S'$ in place of $S$. We find $$\begin{align}\sum_{L\subseteq S}\prod_{x\in L}(x-1)&=\sum_{a\in L\subseteq S}\prod_{x\in L}(x-1)+\sum_{a\notin L\subseteq S}\prod_{x\in L}(x-1)\\ &=\sum_{L\subseteq S'}(a-1)\prod_{x\in L}(x-1)+\sum_{L\subseteq S'}\prod_{x\in L}(x-1)\\ &=a\cdot\sum_{L\subseteq S'}\prod_{x\in L}(x-1)\\ &=a\cdot \prod_{x\in S'}x = \prod_{x\in S}x.\end{align} $$