This is not quite what you're looking for, but here's a theorem (followed by a reference) which justifies the heuristic that the representation theory of a finite group over an algebraically closed field of characteristic $0$ "doesn't depend on the field" :

Suppose $K$ is a field in which every irreducible representation of $G$ is absolutely irreducible. Then for any field extension $K'/K$, the induced morphism on representation rings $R_K(G)\to R_{K'}(G)$ is an isomorphism.

This is in the first paragraph of section 14.6 in Serre's book.

For instance, here's how it can help for statement 1. : if $K$ is an algebraically closed field of characteristic $0$, then

a) all irreducible representations are absolutely irreducible, by Schur's lemma and the existence of eigenvalues

b) $K$ has a common field extension with $\mathbb C$.

From there it's not hard to see that if statement 1. holds over $\mathbb C$, it does so over any algebraically closed field of characteristic $0$ (indeed note that the morphism $R_K(G)\to R_{K'}(G)$ maps the "positive part" to the positive part $R_K^+(G)\to R_{K'}^+(G)$, so if it's an isomorphism so must the latter be; and free commutative monoids have at most one basis - so this induces a bijection between the irreducible representations - apply this to $G_1,G_2$ and $G_1\times G_2$)

You can also deduce 2. from the similar fact over $\overline{\mathbb Q}$ or $\mathbb C$ if you use the other statement from the same paragraph of Serre's book, namely that $R_K(G)\to R_{K'}(G)$ is always injective, no matter what the field extension $K'/K$ is.

Indeed, your statement 2. is then saying that $R_k(S_n)\to R_\overline k(S_n)$ is an isomorphism, but this follows from the following chain of morphisms, where $K$ is a common extension of $\overline k$ and $\mathbb C$: $R_\mathbb Q(S_n)\to R_k(S_n)\to R_\overline k(S_n)\to R_K(S_n)$. By the earlier statement, the last morphism is an isomorphism, but also the composite (if you already know the surjectivity statement for $\mathbb C$), therefore by injectivity so is the middle one.

In other words, the slogan works in these situations, and you can find the appropriate precise statements in this paragraph of Serre's book.

This is not quite what you're looking for, but here's a theorem (followed by a reference) which justifies the heuristic that the representation theory of a finite group over an algebraically closed field of characteristic $0$ "doesn't depend on the field" :

Suppose $K$ is a field in which every irreducible representation of $G$ is absolutely irreducible. Then for any field extension $K'/K$, the induced morphism on representation rings $R_K(G)\to R_{K'}(G)$ is an isomorphism.

This is in the first paragraph of section 14.6 in Serre's book.

For instance, here's how it can help for statement 1. : if $K$ is an algebraically closed field of characteristic $0$, then

a) all irreducible representations are absolutely irreducible, by Schur's lemma and the existence of eigenvalues

b) $K$ has a common field extension with $\mathbb C$.

From there it's not hard to see that if statement 1. holds over $\mathbb C$, it does so over any algebraically closed field of characteristic $0$ (indeed note that the morphism $R_K(G)\to R_{K'}(G)$ maps the "positive part" to the positive part $R_K^+(G)\to R_{K'}^+(G)$, so if it's an isomorphism so must the latter be; and free commutative monoids have at most one basis - so this induces a bijection between the irreducible representations - apply this to $G_1,G_2$ and $G_1\times G_2$)

You can also deduce 2. from the similar fact over $\overline{\mathbb Q}$ or $\mathbb C$ if you use the other statement from the same paragraph of Serre's book, namely that $R_K(G)\to R_{K'}(G)$ is always injective, no matter what the field extension $K'/K$ is.

Indeed, your statement 2. is then saying that $R_k(S_n)\to R_\overline k(S_n)$ is an isomorphism, but this follows from the following chain of morphisms, where $K$ is a common extension of $\overline k$ and $\mathbb C$: $R_\mathbb Q(S_n)\to R_k(S_n)\to R_\overline k(S_n)\to R_K(S_n)$. By the earlier statement, the last morphism is an isomorphism, but also the composite (if you already know the surjectivity statement for $\mathbb C$), therefore by injectivity so is the middle one.

In other words, the slogan works in these situations, and you can find the appropriate precise statements in this first paragraph of section 14.6 of Serre's book.