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 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.

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.

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 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 $(S_{7} \cap C_{2}) \cap A_{10},$ 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.

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 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.