There are multiple way one can construct permutation actions connected to $S$ and $T$:
Approach 1: Consider the wreath product $G \wr H:= G^H \rtimes H$, where $H$ acts on $G^H$ via pre-composition with the right multiplication, i.e. ${^h \eta} := x \mapsto \eta(xh)$. Then $(g,h) \mapsto (\eta_g,h)$ with $\eta_g(x):={^x g}$ is an embedding of $G\rtimes H$ into $G^H \rtimes H$.
Wreath products of permutation groups have two natural permutation actions, on the cartesian product of both and on the set of maps from one set to the other. In our case $G^H\rtimes H$ acts both on $S\times H$ and on $S^H$.
You can "visualise" this like so: Every element of $H$ gives you an automorphism of $G$ and therefore a way to twist (permutation) representations of $G$: For every $h\in H$ we can define $^h S$ as the $G$-set which as a copy of $S$, say $\{^h s \mid s\in S\}$, as elements and the action $({^h g})(^h s) := {^h (gs)}$. With this notation, $S\times H$ is just $\bigsqcup_{h\in H} {^h S}$ and $S^H$ is just $\prod_{h\in H} {^h S}$ where $H$ acts by permuting the ${^h S}$ amongst themselves and $G$ acts on the components individually.
If you like to think in modules: $k[S\times H]$ equals the induced module $\operatorname{Ind}_{G}^{G\rtimes H}(k[S])$ and $k[S\times H]$ equals what is called the "tensor induction" of $k[S]$.
And of course, once we have that we can combine either of those with the quotient $G^H \rtimes H \twoheadrightarrow H$ and the action of $H$ on $T$ to obtain $G^H \rtimes H$-actions on $S\times H \times T$, $(S\times H)\sqcup T$, $S^H\times T$, $S^H \sqcup T$, $(S\times H)^T$, $(S^H)^T$ etc.
Approach 1.5: Or we can use that $H$ not only has the right-multiplication-action on $H$, but also an commuting action by left-multiplication. In particular, $S\times H$ and $S^H$ not only have the $G \wr H$-action, but also a commuting $H$-action. The second construction is therefore to take $(S\times H)\times_H T$ and $S^H\times_H T$ instead of the cartesian products, i.e. to quotient out the additional $H$-action.
The first set is somewhat connected to your guess $S\times T$, but is smaller in general.
Approach 2 is to think of two actions on the same set $\Omega$. If $G$ and $H$ both act on $\Omega$ and the actions commute in the sense that $^g(^h \omega) = {^h({^g\omega})}$, then $G\times H$ acts on $\Omega$. A typical case of this is what's called a "biset", i.e. a set with an left-$G$-action and a compatible right-$H$-action. Think: $G$ acts via left-multiplication and $H$ acts via right-multiplication on $\Omega$ and we want associativity to hold. Actions of the semidirect product are "twisted bisets" in this sense. Think of them as $H$ acting by conjugation and $G$ acting by left-multiplication on $\Omega$.
What's another object on which $G$ acts by left multiplication and $H$ by conjugation? The set $G^T$. $G$ clearly acts by left multiplication in the image and since both $G$ and $T$ are $H$-sets we get a conjugation action of $H$. Together we get the $G\rtimes H$-action $(^{(g,h)}\gamma)(t) := g\cdot{^h\gamma(^{h^{-1}} t)}$.
Similar in the way $S^H$ was similar to $S\times H$ we can also combine the $H$-action on $G\times T$ with the $G$-action by left multiplication (trivial on $T$) to obtain the $G\rtimes H$-action ${^{(g,h)}(x,t)} := (ghxh^{-1},ht)$.
And of course, we can now combine all of these together to get $G\rtimes H$-actions on $G^T \times S^H$ and other funny looking things.
As others have pointed out in the comments, the words should be over the disjoint union of $G$ and $H$. For the sake of completeness, I'm posting an answer to write down what the correct definition would be.
Definition of free product
Let $(G, \cdot_G)$ and $(H, \cdot_H)$ be groups. The free product of $(G, \cdot_G)$ and $(H, \cdot_H)$, denoted $(G, \cdot_G) * (H, \cdot_H)$, is the group $(G * H, \cdot_{G * H})$, where $G * H$ is the set of reduced words on $G$ and $H$, and $\cdot_{G * H}$ is the operation given by concatenation followed by reduction. Precise definitions of each are given below.
The set of reduced words on $G$ and $H$, denoted $G * H$, is defined as $r_{G,H}[W(G \sqcup H)]$, where $W(\cdot)$ denotes the set of words on a set and $r_{G,H}$ is the function that maps each word to its reduced form. That is, it is the image of the set of words on $G \cup H$ under the reduction map $r_{G, H}$.
Let $S$ be a set. The set of words on $S$, denoted $W(S)$, is defined as the set of finite (ordered) tuples in elements of $S$. That is:
$$W(S) := \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{\ge 0},\quad \forall i \in \{1, \ldots, n\}\ x_i \in S \}$$
Thus, in particular:
$$W(G \sqcup H) = \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{\ge 0},\quad \forall i \in \{1, \ldots, n\}\quad (\exists g \in G\ x_i = (g, 1))\ \veebar\ (\exists h \in H\ x_i = (h, 2))\}$$
The reduction map $r_{G,H}$ is the function $r_{G,H} \colon W(G \sqcup H) \to W(G \sqcup H)$ that maps each word to its reduced form. The reduced form of a word is the word that we obtain by repeatedly applying any of the following operations until we can no longer apply any of them.
- Omit a letter that is the identity in $G$ or in $H$ from a word. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, (1_G, 1), x_{i + 1}, \ldots, x_n)$ or $(x_1, \ldots, x_{i - 1}, (1_H, 2), x_{i + 1}, \ldots, x_n)$ into the word $(x_1, \ldots, x_{i - 1}, x_{i + 1}, \ldots, x_n)$.
- Replace two consecutive elements of the same group (either $G$ or $H$) with their product. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, (y_1, \iota), (y_2, \iota), x_{i + 2}, \ldots, x_n)$, where $y_1, y_2$ are both in $G$ or both in $H$ (since $\iota = \iota$) into the word $(x_1, \ldots, x_{i - 1}, (y_1 \cdot y_2, \iota), x_{i + 2}, \ldots, x_n)$, where $\cdot$ denotes correspondingly either $\cdot_G$ or $\cdot_H$.
(I omit the proof that there is a unique reduced form, etc.)
The operation $\cdot_{G * H}$ is defined by concatenating and then reducing. That is, if $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_m)$ are elements of $G * H$, then: $$(x_1, \ldots, x_n) \cdot_{G*H} (y_1, \ldots, y_m) := r_{G,H}\big((x_1, \ldots, x_n, y_1, \ldots, y_n)\big)$$
Proof sketch that it is a group
Closure under the operation. The product of two words is by definition the reduction of their concatenation, so it is also a word in reduced form.
Associativity. The reduction map $r$ is "sub-idempotent", i.e. applying $r$ to any sub-word of a word and then applying it to the entire word is the same as just applying it to the entire word (details omitted). Thus:
$$((x_1, \ldots, x_r) \cdot_{G*H} (y_1, \ldots, y_s)) \cdot_{G*H} (z_1, \ldots, z_t) = r(r((x_1, \ldots, x_r) \mid (y_1, \ldots, y_s)) \mid (z_1, \ldots, z_t)) = r((x_1, \ldots, x_r) \mid (y_1, \ldots, y_s) \mid (z_1, \ldots, z_t)) = r((x_1, \ldots, x_r) \mid r((y_1, \ldots, y_s) \mid (z_1, \ldots, z_t))) = (x_1, \ldots, x_r) \cdot_{G*H} ((y_1, \ldots, y_s) \cdot_{G*H} (z_1, \ldots, z_t))$$
where $\mid$ denotes concatenation.
Identity. The empty word $()$ is the identity, since concatenating with $()$ does nothing, and then reducing doesn't do anything either because the word was already in reduced form.
Inverses. The inverse of $((x_1, \iota_1), \ldots, (x_n, \iota_n))$ is $((x_n^{-1}, \iota_n), \ldots, (x_1^{-1}, \iota_1))$.
Best Answer
If $G$ and $H$ act on $X$ and their actions don't commute, then there's no reason to expect $G \times H$ to act on $X$. Here's an explicit counterexample, followed by a more abstract view that lets you guess the correct answer (so that you can focus on looking for a counterexample rather than a proof).
First, let $G = \mathbb{Z}/2 = \langle \sigma | \sigma^2 \rangle$ and $H = \mathbb{Z}/3 = \langle \tau \mid \tau^3 \rangle$. Then both of these groups act on $X = \{1,2,3\}$, by sending $\sigma$ to the permutation $(1 \ 2)$ and $\tau$ to the permutation $(1 \ 2 \ 3)$. I'll let you check that these really define actions.
Now we ask about $G \times H = \mathbb{Z}/2 \times \mathbb{Z}/3$. It's not hard to check that $G \times H$ has no faithful action on $X$, since $\text{Aut}_X = \mathfrak{S}_3$ (the symmetric group on $3$ letters) has $6$ elements. So $G \times H$ surely doesn't act faithfully. Indeed, by comparing the subgroups of $\mathfrak{S}_3$ to the quotients of $G \times H$ (using the first isomorphism theorem one can show that the only actions${}^1$ of $G \times H$ on $X$ in this case are $(g,h) \cdot x = g \cdot x$ or $(g,h) \cdot x = h \cdot x$! We must simply ignore one of the factors! So there is no action of $G \times H$ that uses the information from both $G$ and $H$.
Now, how could we have seen this coming? An action of $G$ on $X$ can similarly be described as a group homomorphism $G \to \text{Aut}_X$. So your question can be rephrased as "given two homomorphisms $G \to \text{Aut}_X$ and $H \to \text{Aut}_X$ whose images don't commute in $\text{Aut}_X$, is there a group homomorphism $G \times H \to \text{Aut}_X$.
This should ring as false to you. After all, "$G \times H$" is the free way to glue $G$ and $H$ together assuming they commute! In general it's very hard to control maps out of $G \times H$ (here knowing some category theory can also help guide your intuition). So our hope is to search for a counterexample.
The easiest thing to do is to take these maps $G \to \text{Aut}_X$ and $H \to \text{Aut}_X$ to be inclusions, so that we actually have subgroups $G \leq \text{Aut}_X$ and $H \leq \text{Aut}_X$. Then the dream is to find a group $\text{Aut}_X$ with two subgroups $G$ and $H$ (which don't commute in $\text{Aut}_X$) so that $G \times H$ has no interesting homomorphisms into $\text{Aut}_X$. From doing lots of these kinds of problems, I knew that $G = \mathbb{Z}/2$, $H = \mathbb{Z}/3$, and $\text{Aut}_X = \mathfrak{S}_3 = \text{Aut}_{\{1,2,3\}}$ would work. However, this is also one of the first examples you might think to try, since $\mathfrak{S}_3$ is the smallest nonabelian group! So it's the first place a counterexample could possibly be.
As a last aside, I'll say that you can "fix this". Instead of working with the product $G \times H$, you could work with the coproduct (also sometimes called the free product) $G \ast H$. This is the construction that glues two groups together if we don't assume that their images commute! Indeed, it's defined by the property that "for any two homomorphisms $\varphi: G \to K$ and $\psi: H \to K$, there is a unique homomorphism $\varphi \ast \psi : G \ast H \to K$ which "glues $\varphi$ and $\psi$ together".
In this way, if $G$ and $H$ act on $X$ (that is, if we have homomorphisms $G \to \text{Aut}_X$ and $H \to \text{Aut}_X$) then this property guarantees we have a homomorphism $G \ast H \to \text{Aut}_X$ extending the previous homomorphisms. That is, it guarantees an interesting action of $G \ast H$ on $X$ in way that uses the information of the $G$ and $H$ actions.
I hope this helps ^_^
${}^1$: This is a slight fib. We can also have actions like $(g,h) \cdot x = g^{-1} \cdot x$, but these won't exist for other choices of $G$ and $H$, and also won't solve your problem.