(in contrary to what I thought first,) here is a proof that the "exchange condition" holds in the following sense.
It is based on the root configuration in http://arxiv.org/abs/1111.3349 [1].

Let $(W,S)$ be a Coxeter system, $Q \in S^*$ a word in $S$, $w \in W$, $P$ a subword of $Q$. We everywhere consider subwords as being indexed by positions in $Q$, so if $Q = ss$, then the two subwords $s{-}$ und ${-}s$ are considered to be different.

Let $(Q,w,P)$ be the set of subwords $X$ of $Q \setminus P$ such that the **complement** $R = Q \setminus X$ has greedy product $Dem(R) = w$ and the skips in the greedy product are in exactly the positions in $P$, we call these positions the *greedy skips*. (So this is the situation Allen introduced in the question, except that he did not mention the part about the complement.)

Let $D(Q,w,P)$ be the simplicial complex with facets given by $(Q,w,P)$. This complex is clearly pure since every facet contains exactly $len(Q)-\ell_S(w)-len(P)$ many letters.

**Observation 1**: For every facet $X$ of $D(Q,w,P)$, the disjoined union $X \cup (P\setminus P')$ is also a facet of $D(Q,w,P')$ for $P' \subseteq P$.

**Construction 2**: The root configuration in Definition 3.1 in [1] is given for a facet $X$ of $D(Q,w,\{\})$ as follows. Associate to each letter $q_i$ in $Q$ a root $R(X,i)$ by applying the prefix up to position $q_{i-1}$ of the word $Q \setminus X$ of $w$ to the simple root $\alpha_{q_i}$.
In my example $Q = tsstst$, the element $w = sts = tst$ and the facet $t{-}s{--}t$ with complement ${-}s{-}ts{-}$ which is a reduced word for $w$, the root configuration is
$$
R(X,\cdot) =\beta,\alpha,s(\alpha),s(\beta),st(\alpha),sts(\beta)
$$
where $\alpha = \alpha_s, \beta = \alpha_t$ which is equal to
$$
23,12,-12,13,23,-12
$$
where I write $12$ for $\alpha$, $23$ for $\beta$, and $13$ for $\alpha+\beta$.

**Observation 3**: The collections of roots of the root configuration of the complement of $X$ is exactly the inversion set of $w$ (in particular, that's all positive roots).

**Observation 4**: The negative roots in the root configuration could be used as *greedy skips*. This also means that picking a a facet $X$ of $D(Q,w,\{\})$, and a subset $P$ of the letters for which the root configuration is negative results in a facet $X \setminus P$ of $D(Q,w,P)$.

**Observation 5**: If a root in the root configuration is negative, than all appearances of the same root "to the right" are also negative. Analogously, if a root there is positive then all appearances "to the left" are positive.

**Statement 6**: The simplicial complex $D(Q,w,P)$ ~~is vertex-decomposable~~ has the exchange property in the following sense.

Let $q_1$ be the first letter in $Q$
If $\ell_S(q_1w) < \ell_S(w)$, then for any facet $X$ of $D(Q,w,P)$, there exists a facet $Y$ of $D(Q,w,P)$ containing $X \setminus q_1$.

**Proof**:

Consider $X' = X \cup P$ as a facet of $D(Q,w,\{\})$ as in Observation 1.
We now know from Observation 4 that $P$ is a subset of the negative roots in the root configuration of $X'$.
We know from $\ell_S(q_1w) < \ell_S(w)$ that $\alpha_{q_1}$ is in the inversion set of $w$ which is by Observation 3 equal to the root configuration of the complement of $X$.
This provides to obtain $Y'$ by "flipping" $q_1$ in $X'$ to the unique position not in $X'$ for which the root configuration is given by the same simple root $\alpha_{q_1}$.
Call this position $q_i$ and we have $Y' = (X' \setminus q_1 ) \cup q_i$.

It remains to show that $Y'$ contains $P$ and that all roots in the root configuration of $Y'$ at positions in $P$ are negative, since Observation 1 then tells us that $Y = Y' \setminus P$ is the desired facet of $D(Q,w,P)$.

Lemma 3.3(3) in [1] tells us how the root configuration changes when doing the flip $X'$ to $Y'$.
This is indeed easy to see: the roots in the root configuration after $q_i$ are not changed, while we apply $s_{q_1}$ to all the roots before that,
$R(Y',j) = R(X',j)$ for $j>i$, and $R(Y',j) = s_{q_1}\big(R(X',j)\big)$ for $j \leq i$.

Since we only care about the signs, observe that $s_{q_1}$ does only affect the sign of $\alpha_{q_1}$, while all other signs are unchanged.
But since $R(X',i)$ is positive (and ind.eed in equal to $\alpha_{q_1}$), Observations 4 and 5 finally tell us that $P$ is a subset of the letters for which the root configuration of $Y'$ is negative. As desired, we conclude that $Y=Y' \setminus P$ is a facet of $D(Q,w,P)$.