For $n\geq 3$ there exist $x,y\in S_n$ such that $x$ and $y$ have order $2$ and $xy$ has order $n$.
Solution 1:
Take $x$ to be the permutation $i\mapsto n+1-i$, and $z$ to be the forward rotation $i\mapsto i+1$ if $i<n$ and $i\mapsto 1$ if $i=n$. Then clearly $x^2=1$, and $y=xz=z^{-1}x$ has order 2.
Solution 2:
Note that $D_n \leq S_n$ and $D_n$ has the property that you have described.Now write the terms of $D_n$ in permutation notation.
Take $$f= \prod_{i=2}^{\frac{[n+1]}{2}} (i, n+2-i)$$ and $$g=\prod_{i=1}^{\frac{[n]}{2}} (i, n+1-i)$$ then $$fg=(1,2,...,n)$$