It seems to me like this statement is folklore, since e.g. the paper [Stellar Stratifications on Classifying Spaces](https://arxiv.org/abs/1804.11274v2) tries to show a generalization of it and at least hints that my simpler claim is true (see 1.9 there). I think I have found a simple proof, but would be happy about corrections/ comments.

First, note that a regular CW complex $X$ is in particular normal. In particular, the set of cells $S$ carries a partial order where $e_1 \leq e_2$ iff, equivalently,

 - $e_1$ is contained in the closure $\bar{e_2}$,
 - $e_1 \cap \bar{e_2} \neq \emptyset$,

see 3.1 in the mentioned paper for more. To show that the map $\operatorname{Sing}^S (X) \to S$ is an equivalence, we proceed by showing it is essentiall surjective and fully faithful.

**Essentially surjective:** As indicated by the post, from HA A.7.5, we know that the fiber of this map over a cell is just the singular simplicial set of the open cell itself (or, in dimension $0$, a point), in particular contractible and non-empty.

**Fully faithful:** Let $e_1$ and $e_2$ be cells in $X$, and $x \in e_1$, $y \in e_2$. If $e_1 \nleq e_2$, the mapping space $\operatorname{Map}_{\operatorname{Sing}^S(X)}(x,y)$ is also empty since there can't be a path $\gamma : [0,1] \to X$ from $x$ to $y$ that lies over the arrow $e_1 \to e_2$ in $S$, as it would have to somehow jump from $e_1$ to $e_2$ even though $e_1 \cap \bar{e_2} = \emptyset$.

If $e_1 \leq e_2$, so $e_1$ lies in the boundary of $e_2$, we need to show that $\operatorname{Map}_{\operatorname{Sing}^S(X)}(x,y)$ is contractible. As in the proof of HA A.6.10, we can identify this with $\operatorname{Sing}(P)$ with $P$ the space of paths $\gamma: [0,1] \to X$ from $x$ to $y$ such that $\gamma((0,1]) \subseteq e_2$. In particular, $\gamma([0,1]) \subseteq \bar{e_2}$, the image of the gluing map $D^n \to X$ of $e_2$, which by regularity is a homeomorphism onto its image.

Thus, we can identify $P$ with the space of maps $\gamma: [0,1] \to \mathbb{R}^n$ such that $\gamma(0) = y'$ for some fixed $y'$ with $|y'| = 1$ that corresponds to $y$, $\gamma(1) = x'$, and $|\gamma(t)|<1$ for all $0<t\leq 1$. This can clearly be contracted to the linear path, since the open unit ball is convex.