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This conjecture is proven in theorem 7.3 7.4 of "Incidence matrices, geometrical bases, combinatorial prebases and matroids" by T.V. Alekseyevskaya and I.M. Gelfand . See also ($n=2$) and theorem 4.2 4.5 in "Bases in Systems of Simplices and Chambers" by Alekseyevskaya where the same result is proved.for higher $n$.

A previous version of this answer described the similar situation with scissor congruence groups, but I'm no longer certain that one can get a proof using such theorems anymore. It is worth mentioning here that there is a very similar theorem of Pachner which says that two PL-homeomorphic triangulated PL-manifolds are related by a sequence of "bistellar flips", which are essentially the generating relations in your conjecture. I suspect there should be a proof of your conjecture using Pachner's theorem, but I haven't given it much thought.

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This sounds like it could be written down somewhere conjecture is proven 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)theorem 7.3 of "Incidence matrices, geometrical bases, combinatorial prebases 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 matroids" 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 $\tilde{\mathcal P}(X)$T.V. Apparently it was a fact known to Jessen Alekseyevskaya and Thorup, that the relations in $\tilde{\mathcal P}(X)$ are generated by relations of the form you describe (different triangulations are equal)I.M. One reference I found is Gelfand. See also theorem 2.10 4.2 in "Scissor congruences, group homology Bases in Systems of Simplices and characteristic classesChambers" by Dupont (there it Alekseyevskaya where the same result is phrased as $\tilde{\mathcal P}(X)$ being isomorphic to a certain homology group)proved.

Now

A previous version of this answer described the similar situation with scissor congruence groups, but I'm no longer certain that one can get a proof of your conjecture using such theorems anymore. It is worth mentioning here that a relation in $K(S)$ comes from volume forms and can be written as $$\sum \alpha_i \chi_{\Delta_i}\text{dVol}=\sum \beta_i \chi_{\Delta_i}\text{dVol}$$and this implies$$\sum\alpha_i \Delta_i =\sum \beta_i \Delta_i$$in $\tilde{\mathcal P}(X)$ so this relation there is a combination very similar theorem of identities Pachner which says that come from triangulations two PL-homeomorphic triangulated PL-manifolds are related by a sequence of the convex hull on $n+2$ points from the theorem above. Now"bistellar flips", because all the polytopes involved have vertices in $S$, the only triangulations involved which are essentially the ones with vertices in $S$. We conclude that $K(S)$ is generated by identities on simplices as generating relations in your conjecture. I suspect there should be a proof of your conjecture using Pachner's theorem, but I haven't given it much thought.

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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.

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