It's convenient to use the following notations:
$x\sim y\{ Sym(B) \},\quad$if $B$ is a finite nonempty set of numbers, $x, y\in Sym(B)$ and $\exists g \in Sym(B)\ s.t. gxg^{-1}=y$.
$x\sim y\{ Alt(B)\},\quad$if $B$ is a finite nonempty set of numbers, $x, y\in Alt(B)$ and $\exists g \in Alt(B)\ s.t. gxg^{-1}=y$.
Here $Alt(B)$ is the set consisting of all even permutations on the set $B$.
We will use this familiar lemma next:
Lemma:
Let $B$ be a finite nonempty set of numbers, $\forall y \in Alt(B)$, assume the cyclic decomposition of $y$:
$
y= c_1c_2\cdots c_r，$(Counting cycles of length $1$), $ |c_i|=k_i.
\text{If}\ \exists i\ s.t. 2\mid k_i$ or $\exists i\neq j\ s.t.\ k_i=k_j$, then $\forall x \in Alt(B):$
$x\sim y\{ Sym(B) \} \implies x\sim y\{ Alt(B)\}$.
The proof of the lemma we omit.
Back to the original question.
$\forall s\in A_n$,
we want to prove that:($n\ge 5$)
$\begin{align}\\
\exists a,b \in A_n\ :&\ s=aba^{-1}b^{-1}, \iff\\
\exists a,b \in A_n\ :&\ aba^{-1}=sb\iff\\
\exists y\in A_n\ :&\ y\sim sy\{A_n \}\quad (*)
\end{align}
$
Assume the cyclic decomposition of $s$:
$s=c_1c_2\cdots c_r$,
In the following we discuss the three cases in {$c_i$}.
- The number of 3-cycles in {$c_i$} is even.
- The number of 3-cycles in {$c_i$} is odd, $\exists c_i:|c_i|\ge 5$,$|c_i|$ odd.
- The number of 3-cycles in {$c_i$} is odd, $\nexists c_i:|c_i|\ge 5$,$|c_i|$ odd.
First case:
We use the following three steps to process {$c_i$}.
Step $1$:Combine all the cycles of even length in {$c_i$} in any two-by-two combination as:$s_1,s_2,\cdots ,s_p$.At this point each $s_i$ is an even permutation.
Step $2$:Combine all the 3-cycles in {$c_i$} in any two-by-two combination as:$s_{p+1},s_{p+2},\cdots ,s_q$.At this point each $s_i$ is also an even permutation.
Step $3$:The remaining cycles in {$c_i$} are $s_{q+1},s_{q+2},\cdots,s_t$.($2\nmid |s_i|\ge 5$ for $q+1\le i\le t$)
Now $s=s_1s_2\cdots s_t$.$supp\ s_i\cap supp\ s_j = \varnothing$.
We will prove:
$\forall 1\le i\le t,\ \exists\ g_i,\ y_i\in A_n:
g_iy_ig_i^{-1}=s_iy_i\ \& \ supp\ g_i,\ supp\ y_i\ \subseteq supp\
s_i$.
In that case, multiplying these t equations gives:
$(g_1g_2\cdots g_t)(y_1y_2\cdots y_t)(g_1g_2\cdots g_t)^{-1}=s_1s_2\cdots s_ty_1y_2\cdots y_t$.
Therefore(Let $y=y_1y_2\cdots y_t$ in($*$))
$y\sim sy\{A_n\}$.
Consider $s_1$(ditto for $s_2,\cdots,s_p$).
$1.s_1$ is product of two cycles of equal length.
WLOG, let $s_1=(1\ 2\ 3\ \cdots\ 2m)(2m+1\ 2m+2\ \cdots\ 4m)$.(Permutations are multiplied from left to right)
Let $y_1=(4m\quad 4m-1\ \cdots\ 2m+2\quad 1\quad 2m+1),\ |y_1|=2m+1\ \implies$
$s_1y_1=(1\ 2\ \cdots\ 2m+1),\ |s_1y_1|=2m+1$.
So $y_1\sim s_1y_1\{S_{4m}\}\implies y_1\sim s_1y_1\{A_{4m}\}$ (by the lemma)
$\implies \exists g_1,\ y_1\in A_n:g_1y_1g_1^{-1}=s_1y_1\ \&\ supp\ g_1,\ supp\ y_1\subseteq supp\ s_1=[4m]$.
$2.s_1$ is product of two cycles of distinct length.
WLOG, let $s_1=(1\quad 2\quad \cdots\quad 2v+2m)(2v+2m+1\quad 2v+2m+2\quad \cdots \quad 2v+4m)$.
$a=(1\quad 2\ \cdots\ 2v+2m),\ b=(2v+2m+1\quad 2v+2m+2\ \cdots \ 2v+4m)$.
Let $y_1=b^{-1}c,\ $where $c=(1\quad 2v+1\quad 2\quad 4\quad 5\quad 6\quad \cdots\quad 2v)\\
s_1y_1=ac=(1\quad 4\quad 6\quad 8\quad \cdots\quad 2v\quad 2\quad 3\quad 5\quad 7\quad \cdots\quad 2v-1)(2v+1\quad 2v+2\quad \cdots\quad 2v+2m)
$.
$y_1\ \&\ s_1y_1$ are both product of disjoint $2v$-cycle and $2m$-cycle.
So $y_1\sim s_1y_1\{S_{2v+4m}\}\implies y_1\sim s_1y_1\{A_{2v+4m}\}$ (by the lemma)
$\implies \exists g_1,\ y_1\in A_n:g_1y_1g_1^{-1}=s_1y_1\ \&\ supp\ g_1,\ supp\ y_1\subseteq supp\ s_1=[2v+4m]$.
Consider $s_{p+1}$(ditto for $s_{p+2},\cdots,s_q$).
WLOG, let $s_{p+1}=(1\quad 2\quad 3)(4\quad\ 5\quad 6)$.
Let $y_{p+1}=s_{p+1},\ g_{p+1}=(1\quad 2)(4\quad 5)\implies g_{p+1}y_{p+1}g_{p+1}^{-1}=(1\quad 3\quad 2)(4\quad 6\quad 5)=y_{p+1}s_{p+1}$.
Consider $s_{q+1}$(ditto for $s_{p+2},\cdots,s_t$).
$1.|s_{q+1}|=5$
WLOG, $S_{q+1}=(1\quad 2\quad 3\quad 4\quad 5)$,
let $y_{q+1}=(1\quad 3\quad 2)\implies s_{q+1}y_{q+1}=(3\quad 4\quad 5)\implies y_{q+1}\sim s_{q+1}y_{q+1}\{A_5\}$.
$2.|s_{q+1}|=2k+1,\ k\ge 3$
WLOG, $s_{q+1}=(1\quad 2\quad 3\quad \cdots\quad 2k+1)$,
let $y_{q+1}=(1\quad 6\quad 7\quad \cdots\quad 2k+1\quad 3\quad 2)\quad (|y_{q+1}|=2k-1)\implies\\ s_{q+1}y_{q+1}=(3\quad 4\quad 5\quad 7\quad 9\quad \cdots\quad 2k+1\quad 6\quad 8\quad \cdots\quad 2k)\implies y_{q+1}\sim s_{q+1}y_{q+1}\{A_{2k+1}\}$.
Thus we have proved First case.
The remaining two cases can be proved similarly.
