If an arbitrary permutation group $P$ on a set $X\cup Y$ with $X\cap Y=\emptyset$ has two subgroups $G,H\subseteq P$ where $G$ is a permutation group on $X$ and $H$ is a permutation group on $Y$ satisfying $P\cong G\times H$ where $\times$ is the group direct product. Is it always true we must have:
$$P=\{\sigma\cup \phi\in \text{Sym}(X\cup Y):(\sigma,\phi)\in G\times H\}$$
Where $\sigma\cup \phi$ is the standard union of binary relations interpreting $\sigma$ and $\phi$ as functionals relations. So that $(\sigma\cup \phi)(t)=\sigma(t)$ if $t\in X$ and $(\sigma\cup \phi)(t)=\phi(t)$ if $t\in Y$. I mean I think it it is. At the very least its clearly true for finite groups since $\sigma\in G\land \phi\in H\implies \sigma\cup \phi\in P$ therefore we have $\{\sigma\cup \phi\in \text{Sym}(X\cup Y):(\sigma,\phi)\in G\times H\}\subseteq P$ and since we assumed the group $P$ was finite that $|\{\sigma\cup \phi\in \text{Sym}(X\cup Y):(\sigma,\phi)\in G\times H\}|=|G\times H|=|P|$ which means they must be equal. However what about the cases when $P$ isn't a finite group? Also if this is true, then why would anyone bother using the group direct product to express the isomorphism between products of permutation subgroups when they could write the definition out explicitly by the above identity? I also get the feeling this might be rather trivial, in which case I apologies for wasting anyone's time.
This is false in general, because a group $F$ can be isomorphic to a proper subgroup $G$ of itself. For example you could have $K = \mathbb Z $ and $G = 2{\mathbb Z}$.
Then $P$ could have subgroups $G$ and $H$ as described (where $H$ can be any group), but $P=\{\sigma \cup \phi \in {\rm Sym}(X \cup Y):(σ,ϕ) \in F \times H\}$. So we have $P \cong G \times H$, but your equality is not true with $G$ in place of $F$.
OK, let's be more specific. We take $X = {\mathbb Z}$, $Y = \{y \}$ a set with a single element not in $X$, $F = \langle \sigma \rangle$ with $\sigma:n \mapsto n+1$ for $n \in X $, $G = \langle \sigma^2 \rangle$, $H$ the trivial group, and $P=\{\sigma \cup \phi \in {\rm Sym}(X \cup Y):(σ,ϕ) \in F \times H\}$.
Then $P \cong G \times H$ (an infinite cyclic group), but $P \ne \{\sigma \cup \phi \in {\rm Sym}(X \cup Y):(σ,ϕ) \in G \times H\}$.