A complete combinatorial proof using Allen's comment:
Let $(W,S)$ be a Coxeter system, and let $Dem(T) \in W$ be the Demazure product or greedy product of a word $T$ in $S$.
Claim 1: $Dem(T)$ is the unique Bruhat maximal element in $\big\{ \prod Q : Q \subseteq T\big\}$.
The idea is to start with a subword $Q$ of $T$ and compare it with the subword $D$ of $T$ picked by the greedy product. You scan through $Q$ from left to right and if you see a letter that is picked in $D$ but not in $Q$, you insert it into $Q$. If this goes up in Bruhat order, we are fine in doing so, and if you go down in Bruhat order, you find by the exchange condition a letter to its right that you can remove in exchange for the inserted letter. By this procedure, you only go up in Bruhat order and we are done.
And now to the formal proof: This is a consequence of the following lifting property in Bruhat order as described in Proposition 2.2.7 of Björner-Brenti's Combinatorics of Coxeter groups
Lemma 1 (lifting property): Let $u < w$ in Bruhat order, and let $s$ be a right descent of $w$ but not of $u$. Then $us < w$.
(Proof in Björner-Brenti.)
Lemma 2: Let $u \leq w$ in Bruhat order, and let $s$ be a right ascent of $w$. Then $us \leq ws$.
Proof: Since $u \leq w$, we have that a reduced expression $a$ for $u$ which is a subword of a reduced expression $b$ for $w$. But since now $bs$ is a reduced expression for $ws$, it contains the expression $as$ (which might or might not be reduced) and we are done. $\square$
Proof of claim by induction: Let $T = t_1\cdots t_m$. The case $m \in \{0,1\}$ is trivial, so assume $m>1$, let $T' = t_1\cdots t_{m-1}$ and we know that $Dem(T')$ is the unique Bruhat maximal element in $\{ \prod Q : Q \subseteq T'\}$.
Let $Q$ be a subword of $T$. If $Q$ is a subword of $T'$ we are done since by assumption $\prod Q < Dem(T') \leq Dem(T)$, so we only treat the case that $Q$ uses the last letter $t_m$.
We have $Q \setminus t_m$ is a subword of $T'$ so $\prod Q\setminus t_m \leq Dem(T')$ by induction.
If $Dem(T) > Dem(T')$, we are in the situation of Lemma 2 and conclude $$\prod Q \leq Dem(T') \cdot t_m = Dem(T).$$
If $Dem(T) = Dem(T')$, we are in the situation of Lemma 1 and conclude $$\prod Q < Dem(T') = Dem(T). \quad \square$$
Corollary: The Demazure product is well-defined in Artin groups. This is, let $T$ be a word of $S$ and let $T'$ be obtained from $T$ by a braid move. Then $Dem(T) = Dem(T')$.
Proof: $T'$ is obtained from $T$ by replace a consecutive substring $x = stst\ldots$ of length $m(s,t)$ by $y = tsts\ldots$. For any subword $Q$ of $T$, one can now choose the same subword in $T′$ *as long as $Q$ does not contain all of $x$. But if this is the case, one can choose the subword $Q'$ of $T'$ where $Q'$ is obtained from $Q$ by using $y$ instead of $x$. $\square$
(As usual with MO proofs, please let me know if something is unclear or plainly wrong.)