Does the symmetric group $S_{10}$ factor as a knit product of symmetric subgroups $S_6$ and $S_7$?

Here are a few comments and a slightly different approach, though we take advantage of some of the earlier comments. We first note the well-known (at least to people who work with factorizations) fact that if the finite group $G$ has a factorization of the form $G = AB$ with $A \cap B = 1 $ and $A,B$ subgroups, then we have $A \cap B^{g} = 1$ for all $g \in G$- for we also have $G = BA,$ and if we write $g = ba$ for some $b \in B, a \in A,$ then we have $A \cap B^{g} = A \cap B^{a} = (A \cap B)^{a} = 1.$

This means that if $G = S_{10}$ has a factorization of the form $G = AB$ with $A \cong S_{6}$ and $B \cong S_{7}$ (so that $A \cap B = 1$ as noted in the body of the question), then no non-identity element of $A$ can have the same disjoint cycle structure (in the given embedding) as any non-identity element of $B.$

Now $S_{6}$ contains commuting (and conjugate) distinct involutions which are odd permutations (in the natural representation). Hence the subgroup $A$ above contains an involution which is an even permutation in the embedding in $S_{10}.$ This is either a product of two disjoint transpositions or product of four disjoint transpositions.

It is noted in comments that $B$ must be of the form $(S_{7} \times C_{2}) \cap A_{10},$ where the $S_{7}$ is a "natural" $S_{7}$ inside $S_{10},$ ie fixing three points. It follows that $B$ contains both involutions which are products of two disjoint transpositions, and involutions which are product of four disjoint transpositions, contrary to the remarks above.

As has been pointed out in comments, the only subgroups $G$ and $H$ of $S_{10}$ isomorphic to $S_7$ and $S_6$ that could possibly have trivial intersections are the copy of $S_7$ that lies in $A_{10}$ and the copy of $S_6$ that is transitive on the $10$ points.

Note that $S_6 \cong {\rm P \Sigma L}(2,9)$, the extension of ${\rm PSL}(2,9)$ by a field automorphism, and the transitive $S_6$ in $S_{10}$ can be described that way.

There are only $360$ conjugates of $G \cong S_7$ in $S_{10}$, so it is an easy computer calcualtion to check that they all have intersection of order $2$ with ${\rm P \Sigma L}(2,9)$: see Magma calcualtion below. So there is no such factorization.

> G := sub<Sym(10) | (1,2,3), (1,2,3,4,5,6,7), (1,2)(3,4)(5,6)(8,9)>;
> isiso := IsIsomorphic(G,Sym(7)); isiso;                            
true
> H := PSigmaL(2,9);
> isiso := IsIsomorphic(H,Sym(6)); isiso;
true
> N := Normaliser(Sym(10),G);
> Index(Sym(10),N);
360
> T := Transversal(Sym(10),N);
> forall{t: t in T | #(G^t meet H) eq 2};
true