Set $M = \Spec(C)$ and $M' = \Spec(C')$. Write $U = \Spec(A)$, $U' = \Spec(A')$, $R = \Spec(B)$, and $R' = \Spec(B')$. We will use the results of Lemmas \ref{lemma-integral-over-invariants}, \ref{lemma-invariants-base-change}, and \ref{lemma-points} without further mention. \medskip\noindent Assume $C$ is a strictly henselian local ring. Let $p \in M$ be the closed point and let $p' \in M'$ map to $p$. Claim: in this case there is a disjoint union decomposition $(U', R', s', t', c') = (U, R, s, t, c) \amalg (U'', R'', s'', t'', c'')$ over $(U, R, s, t, c)$ such that for the corresponding disjoint union decomposition $M' = M \amalg M''$ over $M$ the point $p'$ corresponds to $p \in M$. \medskip\noindent The claim implies the lemma. Suppose that $M_1 \to M$ is a flat morphism of affine schemes. Then we can base change everything to $M_1$ without affecting the hypotheses (1) -- (4). From Lemma \ref{lemma-invariants-base-change} we see $M_1$, resp.\ $M_1'$ is the spectrum of the $R_1$-invariant functions on $U_1$, resp.\ the $R'_1$-invariant functions on $U'_1$. Suppose that $p' \in M'$ maps to $p \in M$. Let $M_1$ be the spectrum of the strict henselization of $\mathcal{O}_{M, p}$ with closed point $p_1 \in M_1$. Choose a point $p'_1 \in M'_1$ mapping to $p_1$ and $p'$. From the claim we get $$ (U'_1, R'_1, s'_1, t'_1, c'_1) = (U_1, R_1, s_1, t_1, c_1) \amalg (U''_1, R''_1, s''_1, t''_1, c''_1) $$ and correspondingly $M'_1 = M_1 \amalg M''_1$ as a scheme over $M_1$. Write $M_1 = \Spec(C_1)$ and write $C_1 = \colim C_i$ as a filtered colimit of \'etale $C$-algebras. Set $M_i = \Spec(C_i)$. The $M_1 = \lim M_i$ and similarly for the other schemes. By Limits, Lemmas \ref{limits-lemma-descend-opens} and \ref{limits-lemma-descend-isomorphism} we can find an $i$ such that $$ (U'_i, R'_i, s'_i, t'_i, c'_i) = (U_i, R_i, s_i, t_i, c_i) \amalg (U''_i, R''_i, s''_i, t''_i, c''_i) $$ We conclude that $M'_i = M_i \amalg M''_i$. In particular $M' \to M$ becomes \'etale at a point over $p'$ after an \'etale base change. This implies that $M' \to M$ is \'etale at $p'$ (for example by Morphisms, Lemma \ref{morphisms-lemma-set-points-where-fibres-etale}). We will prove $U' \cong M' \times_M U$ after we prove the claim. \medskip\noindent Proof of the claim. Observe that $U_p$ and $U'_{p'}$ have finitely many points. For $u \in U_p$ we have $\kappa(u)/\kappa(p)$ is algebraic, hence $\kappa(u)$ is separably closed. As $U' \to U$ is \'etale, we conclude the morphism $U'_{p'} \to U_p$ induces isomorphisms on residue field extensions. Let $u' \in U'_{p'}$ with image $u \in U_p$. By assumption (3) the morphism of scheme theoretic fibres $(s')^{-1}(u') \to s^{-1}(u)$, $(t')^{-1}(u') \to t^{-1}(u)$, and $G'_{u'} \to G_u$ are isomorphisms. Observing that $U_p = t(s^{-1}(u))$ (set theoretically) we conclude that the points of $U'_{p'}$ surject onto the points of $U_p$. Suppose that $u'_1$ and $u'_2$ are points of $U'_{p'}$ mapping to the same point $u$ of $U_p$. Then there exists a point $r' \in R'_{p'}$ with $s'(r') = u'_1$ and $t'(r') = u'_2$. Consider the two towers of fields $$ \kappa(r')/\kappa(u'_1)/\kappa(u)/\kappa(p) \quad \kappa(r')/\kappa(u'_2)/\kappa(u)/\kappa(p) $$ whose ``ends'' are the same as the two ``ends'' of the two towers $$ \kappa(r')/\kappa(u'_1)/\kappa(p')/\kappa(p) \quad \kappa(r')/\kappa(u'_2)/\kappa(p')/\kappa(p) $$ These two induce the same maps $\kappa(p') \to \kappa(r')$ as $(U'_{p'}, R'_{p'}, s', t', c')$ is a groupoid over $p'$. Since $\kappa(u)/\kappa(p)$ is purely inseparable, we conclude that the two induced maps $\kappa(u) \to \kappa(r')$ are the same. Therefore $r'$ maps to a point of the fibre $G_u$. By assumption (3) we conclude that $r' \in (G')_{u'_1}$. Namely, we may think of $G$ as a closed subscheme of $R$ viewed as a scheme over $U$ via $s$ and use that the base change to $U'$ gives $G' \subset R'$. In particular we have $u'_1 = u'_2$. We conclude that $U'_{p'} \to U_p$ is a bijective map on points inducing isomorphisms on residue fields. It follows that $U'_{p'}$ is a finite set of closed points (Algebra, Lemma \ref{algebra-lemma-finite-residue-extension-closed}) and hence $U'_{p'}$ is closed in $U'$. Let $J' \subset A'$ be the radical ideal cutting out $U'_{p'}$ set theoretically. \medskip\noindent Second part proof of the claim. Let $\mathfrak m \subset C$ be the maximal ideal. Observe that $(A, \mathfrak m A)$ is a henselian pair by More on Algebra, Lemma \ref{more-algebra-lemma-integral-over-henselian-pair}. Let $J = \sqrt{\mathfrak m A}$. Then $(A, J)$ is a henselian pair (More on Algebra, Lemma \ref{more-algebra-lemma-change-ideal-henselian-pair}) and the \'etale ring map $A \to A'$ induces an isomorphism $A/J \to A'/J'$ by our deliberations above. We conclude that $A' = A \times A''$ by More on Algebra, Lemma \ref{more-algebra-lemma-characterize-henselian-pair}. Consider the corresponding disjoint union decomposition $U' = U \amalg U''$. The open $(s')^{-1}(U)$ is the set of points of $R'$ specializing to a point of $R'_{p'}$. Similarly for $(t')^{-1}(U)$. Similarly we have $(s')^{-1}(U'') = (t')^{-1}(U'')$ as this is the set of points which do not specialize to $R'_{p'}$. Hence we obtain a disjoint union decomposition $$ (U', R', s', t', c') = (U, R, s, t, c) \amalg (U'', R'', s'', t'', c'') $$ This immediately gives $M' = M \amalg M''$ and the proof of the claim is complete. \medskip\noindent We still have to prove that the canonical map $U' \to M' \times_M U$ is an isomorphism. It is an \'etale morphism (Morphisms, Lemma \ref{morphisms-lemma-etale-permanence}). On the other hand, by base changing to strictly henselian local rings (as in the third paragraph of the proof) and using the bijectivity $U'_{p'} \to U_p$ established in the course of the proof of the claim, we see that $U' \to M' \times_M U$ is universally bijective (some details omitted). However, a universally bijective \'etale morphism is an isomorphism (Descent, Lemma \ref{descent-lemma-universally-injective-etale-open-immersion}) and the proof is complete.