If the Galois group is $S_3$, can the extension be realized as the splitting field of a cubic?

We know that the Galois group of an irreducible cubic polynomial is $S_3$ or $A_3$, but is every group extension whose Galois group is $S_3$ a splitting field of a cubic polynomial? If not, the extension must be the splitting field of a polynomial of degree 6. And therefore $S_3$ must be a transitive subgroup of $S_6$. Unfortunately, I found (1 2 3)(4 5 6) and (1 4)(3 5)(2 6) can generate such transitive $S_3$, but I can't find the corresponding polynomial.

Is it true that every Galois extension with Galois group $S_3$ a splitting field of a cubic irreducible polynomial? Thanks for your help!

You have two questions (since edited to fix the subject line, but I'll keep both answers): one in the subject line: "is there an irreducible polynomial that is not cubic whose Galois group is $S_3$?"; and one in the post: "is every Galois extension (of $\mathbb{Q}$?) with Galois group $S_3$ a splitting field of a cubic irreducible polynomial?"

By Steinitz's Theorem, a finite extension $F/E$ is simple (that is, there exists $a\in F$ such that $F=E(a)$) if and only if there are only finitely many intermediate fields $K$, $E\leq K\leq F$. This is the case in a finite Galois extension, by the correspondence theorem.

Question 1. If $F$ is Galois over $E$ and has Galois group $S_3$ over $E$, then the extension is simple by Steinitz's Theorem, so there exists $a\in F$ such that $F=E(a)$. In particular, the degree of the minimal irreducible polynomial of $a$ over $E$ is $[F:E]=6$, and its splitting field is $F$, because $F$ is Galois so any irreducible polynomial in $E[x]$ either is irreducible or splits over $F$. Thus, $F$ is the splitting field of an irreducible cubic.

Question 2. Suppose $F$ is Galois over $E$ with $\mathrm{Gal}(F/E)\cong S_3$. By the correspondence theorem, a subgroup of order $2$ of $S_3$ corresponds to an intermediate extension $K$, $E\leq K\leq F$, with $[F:K]=2$ (and therefore $[K:E]=3$); and $K$ is not Galois over $E$ because the subgroup of order $2$ is not normal in $S_3$. Let $a\in K$ be a primitive element of $K$ over $E$. Since $[K:E]=3$, the irreducible polynomial of $a$ over $E$ is a cubic; its splitting field is Galois over $E$ and contains $K$, and thus must equal $F$. So $F$ is the splitting field of the monic irreducible of $a$ over $E$, which is a cubic. Thus, all such extensions arise as splitting fields of irreducible cubics.

If $K/\mathbb{Q}$ has Galois group $S_3$, then because $S_3$ has a subgroup of index 3, this corresponds to a subextension $F$ of degree 3. A primitive element for $F$ will be the root of a cubic polynomial and its splitting field will be the original $K$ (it can't be $F$ itself because $S_3$ has no normal subgroups of index 3).

Of course, you can also find a degree 6 polynomial for which $K$ is the splitting field (see Arturo Magidin's comment).