This sounds like it could be written down somewhere in the literature on Hilbert's third problem. I believe I can sketch a proof.
Let $X$ be a space (which will be $\mathbb R^n$ for us), and $G$ a group of isometries of $X$. We define $\mathcal P(X,G)$ to be the $G$-scissor congruence group of $X$. It is defined as a subgroup of the free abelian group generated by polytopes in $X$, modulo the following relations.
$\tag{1} P=\sqcup P_i \implies [P]=\sum [P_i]$
$\tag{2} [P]=[gP], \forall g\in G$
So two polytopes are equivalent if they can be decomposed into pieces which are equivalent modulo $G$. When $G=\lbrace 0\rbrace$ let's call this group $\tilde{\mathcal P}(X)$. Apparently it was a fact known to Jessen and Thorup, that the relations in $\tilde{\mathcal P}(X)$ are generated by relations of the form you describe (different triangulations are equal). One reference I found is theorem 2.10 in "Scissor congruences, group homology and characteristic classes" by Dupont (there it is phrased as $\tilde{\mathcal P}(X)$ being isomorphic to a certain homology group).
Now a proof of your conjecture is that a relation in $K(S)$ coming comes from volume forms and can be written as
an equivalence of linear combinations of polytopes, $$\sum \alpha_i \chi_{\Delta_i}\text{dVol}=\sum \beta_i \chi_{\Delta_i}\text{dVol}$$
and this lives implies
$$\sum\alpha_i \Delta_i =\sum \beta_i \Delta_i$$
in $\tilde{\mathcal P}(X)$ so this relation is a combination of identities that come from triangulations of the convex hull on $n+2$ points from the theorem above. Now, because all the polytopes involved have vertices in $S$, the only triangulations involved are the ones with vertices in $S$. We conclude that $K(S)$ is generated by identities on simplices as in your conjecture.
