Let $A_{\Gamma'}$ be the group asoociated to any graph $\Gamma'$. Concerning your first question I can show that there is a short exact sequence $$1\rightarrow F \rightarrow A_\Gamma \rightarrow A_{\Gamma \setminus e}\rightarrow 1.$$$$1\rightarrow F \rightarrow A_{\Gamma \setminus e}\rightarrow A_\Gamma\rightarrow 1.$$
Let me work in the case of graph products (which includes both the right angled Artin and the right andgled Coxeter cases).
We obtain a presentation for $A_\Gamma$ by adding a relation to a presentation for $A_{\Gamma\setminus e}$.
So there is a canonical group homomorphism from $A_{\Gamma\setminus e}$ to $A_\Gamma$. So maybe you are wondering what the Kernel looks like.
Instead of answering this question let me instead consider the homomorphism $A_{\Gamma}\rightarrow A_{\Gamma\setminus v}$$A_{\Gamma\setminus e}\rightarrow A_{\Gamma\setminus v}$ where $v$ is one of the endpoints of $e$ (Of course we also have to remove all edge to $v$). By the argument of Section 4 in Holt,Rees its Kernel is a free product of copies of the vertex group. We have a factorization $A_{\Gamma}\rightarrow A_{\Gamma\setminus e}\rightarrow A_{\Gamma\setminus v}$$A_{\Gamma\setminus e}\rightarrow A_\Gamma \rightarrow A_{\Gamma\setminus v}$. So especially the Kernel of the map we were originally interested in is contained in that Kernel.
In the Artin case we get immediately that this kernel is free (as it is a subgroup of a free group) and in the Coxeter case it is also a free group. To show this it is enough to show that it is torsionfree (using the action of a free product on a tree). But this is also clear since any torsion element in a free product is conjugate to an element in one of the factor groups. Such elements cannot lie in the Kernel of $A_{\Gamma\setminus e}$ to $A_\Gamma$ (but to show this I would have to make the isomorphism from the kernel to such a free product explicit. It is in the paper mentioned above but writing it down would take quite some notation).