Problem: If $H\subset F$ is another $\sigma$-algebra and $\sigma(X,G)=\sigma(\sigma(X)\cup G)$ is independent of $H$ then $\mathbb{E}(X|\sigma(G,H))=E(X|G)$
Attempt: We want to show that $$\mathbb{E}(X|\sigma(G,H))=\mathbb{E}(X|\sigma(G))$$ which, in other words, means that we should show that any version of the left-hand side is a version of the right-hand side and vice-versa.
Let us define the set $S=\{A\subseteq \sigma(G,H)| \mathbb{E}(\mathbb{E}(X|\sigma(G))1_{[A]})=\mathbb{E}(\mathbb{E}(X|\sigma(G,H))1_{[A]})\}$. First, we show that $S$ contains both the elements of $\sigma(G)$ (by definition) and of $\sigma(H)$ (by independence) and is closed under intersection. Through arguments mainly by independence of $H$, it remains to show that $S$ is also closed under unions and complements.
Can somebody please provide guidance as to whether or not this approach is sound, and if there is anything noteworthy to keep under consideration for this problem?
This approach seems okay, but people typically invoke a monotone class theorem or Dynkin's $\pi$-$\lambda$ theorem to do the heavy lifting.
The set $K=\{A\cap B : A\in G, B\in H\}$ is a $\pi$-system (i.e., closed under finite intersections). Your set $S$ is a $\lambda$-system: it includes $\Omega$, and it is closed under complements and countable disjoint unions (by linearity of the expectation and elementary properties of indicator variables). If you show that $K\subseteq S$, then $\sigma(K)\subseteq S$ by the $\pi$-$\lambda$ theorem. And since $G \cup H \subseteq K$, we must have $\sigma(G\cup H)\subseteq \sigma(K)\subseteq S$.