In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.
Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?
After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S_\bullet/R} \otimes_{S_\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.
Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.
P.S. I don't understand why Scholze uses $S_\bullet$ in the definition of $\Bbb L_{S/R}$ when he uses $S^\bullet$. Indeed, Gabber–Romero uses $P^\bullet$ in Lemma 6.5.9.
Best Answer
Regarding the first and third question, what you say is correct. For the second question, you are looking for the base change compatibility of the cotangent complex: If $R\to R'$ is any map of rings and $S$ is an $R$-algebra such that $S'=S\otimes^L_R R'$ sits in degree $0$, then $$R'\otimes^L_R \mathbb L_{S/R}\cong \mathbb L_{S'/R'}.$$
To see this, use any simplicial resolution $S_\bullet$ of $S$ by free $R$-algebras; then $S_\bullet\otimes_R R'$ is such a resolution of $S'$. The left-hand side is computed by the simplicial $R'$-module $R'\otimes_R(\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S)$, and the right-hand side by $\Omega^1_{S'_\bullet/R'}\otimes_{S'_\bullet} S'$; these two simplicial $R'$-modules agree.
Implicit here is that one can use any free resolution to compute the cotangent complex, not necessarily the standard functorial one.
But I think there must be some more basic confusion on cotangent complexes, as you say that you do not see the relation between $\mathbb L_{S/R}$ and $\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S$. To address this: What do you take as the definition of the cotangent complex?