Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic? If $f(x)$ is an irreducible cubic, then $\operatorname{Gal}(f(x))\cong S_3$ or $A_3$. But what about the converse? That is, if $\operatorname{Gal}(K/F)\cong S_3$, is it necessarily true that $K$ is the splitting field of some irreducible cubic in $F[x]$?
Yes. To see this, note that if $K/F$ is a Galois extension, then by the Fundamental Theorem of Galois Theory there is an intermediate subfield $L$ with $[L : F] = 3$, corresponding to the fixed field of $\langle 
(1,2) \rangle$. By the Primitive element theorem, which in particular holds for any Galois extension, $L = F[\theta]$ for some $\theta \in L$. Let $p(x)$ denote the minimal polynomial of $\theta$. Then $degree(p(x)) = 3$. 

Now let $r \in K$ be another root of $p(x)$. If $p \in L$, then $L/F$ would also be a Galois extension--which contradicts the fact that the group $\langle 
(1,2) \rangle $ is not a normal subgroup of $S_3$. Thus claim $r \notin L$, which gives that $K$ is the splitting field of $p(x)$.
A quick addition to the post that I thought will be useful for someone else who need a little more clarification:  

$K$ being a Galois extension implies that it is Separable.
Since $p(x)$ has a root, $\theta$ in $L\subset K$, it says that $p(x)$ should linearly split in $K$ (By definition of  separability ).
Notice that for a splitting field, having all the roots is not enough. It should also be generated by those roots

Now as it is proved that other remaining two roots cannot be in the intermediate field, and from Galois correspondence as we know that there are no intermediate fields between $L$ and $K$, it must have that $F[\theta,r,s]=K$