Does Koszul duality between $Comm$ and $Lie$ imply the power series identity $\exp(\ln(1-z))-1 = -z$?

Let me flesh out the answer a little. The general statement is given by Theorem 7.5.1 in the book Algebraic Operads by Loday and Vallette.

First a definition. Let $P = P(E,R)$ be a quadratic operad, with generators $E$ (f.gen. s.t. $E(0) = 0$) and $R \subset E \circ E$ quadratic relations. Let $P^{(r)}(n)$ be the subspace of operations of weight $r$, where $E$ is of weight $1$. There is a generating series, aka Hilbert-Poincaré series: $$f^P(x,y) = \sum_{r \ge 0, n \ge 1} \frac{\dim P^{(r)}(n)}{n!} y^r x^n.$$

The theorem states that if $P$ is Koszul, with dual $P^!$, then there is a functional equation: $$f^{P^!}(f^P(x,y),-y) = x.$$

Remark: as Nicholas Kuhn explained, this equality follows from the acyclicity of the Koszul complex, the product $P^¡ \circ P$ with the Koszul differential.

$\newcommand{\Com}{\mathsf{Com}}\newcommand{\Lie}{\mathsf{Lie}}$ Apply this to $P = \Lie$, $P^! = \Com$. It's well-known that $\Com(n) = \Com^{(n-1)}(n)$ is of dimension $1$ for $n \ge 1$, while $\Lie(n) = \Lie^{(n-1)}(n)$ is of dimension $(n-1)!$ for $n \ge 1$. So in particular you get \begin{align} f^\Com(x,1) & = \sum_{n \ge 1} \frac{x^n}{n!} = \exp(x) - 1,\\ f^\Lie(x,-1) & = \sum_{n \ge 1} \frac{(-1)^{n-1} x^n}{n} = \ln(1+x) \end{align}

Apply the functional equation to $y = -1$ and you get $\exp(\ln(1-x))-1=x$.

Yes. (Mathoverflow won't let me make this my total answer, so ...)

Koszul duality says that a certain chain complex of graded vector spaces is acyclic. Thus the alternating sum of the Poincare series gives $z$.