I think the problem is simple but I just want to make sure I am doing it right.
Elements of order $7$ are of the form $(abcdefg)$. There are $6!$ such elements. But $6! \nmid 7!/2$ and thus if it was a conjugacy class, it would contradict orbit stabilizer theorem.
Yes, your solution is good.
In fact, it is not too hard to determine a "splitting" criterion (if you already understand wreath products, anyway).
Permutations of a given cycle type are a conjugacy class in $S_n$. As an $S_n$-set, by Orbit-Stabilizer the conjugacy class of $\sigma$ is equivalent to the coset space $S_n/C(\sigma)$. Under conjugation, the $A_n$-orbit of $\sigma$ corresponds to the subset $A_nC(\sigma)/C(\sigma)$ of $S_n/C(\sigma)$. The conjugacy class "inerts" (doesn't split) when
$$ \left|\frac{S_n}{C(\sigma)}\right|=\left|\frac{A_nC(\sigma)}{C(\sigma)}\right|=\left|\frac{A_n}{A_n\cap C(\sigma)}\right| $$
which may be rearranged to
$$ 2=\left|\frac{S_n}{A_n}\right|=\left|\frac{C(\sigma)}{A_n\cap C(\sigma)}\right|. $$
The sign homomorphism $C(\sigma)\to\mathbb{Z}_2$ shows $C(\sigma)$ is either half even / half odd elements (if the homomorphism is onto) or all even elements (if the homomorphism is trivial). Thus, the only time the equality above fails (i.e. the conjugacy class of $\sigma$ splits in $A_n$) is if $C(\sigma)\subseteq A_n$ is all even elements.
Let $c_k(\sigma)$ denote the number of $k$-cycles of $\sigma$. Then
$$ C(\sigma)=\prod_{k=1}^n \mathbb{Z}_k\wr S_{c_k(\sigma)} $$
is a direct product of wreath products of cyclic groups. The $c_k(\sigma)$-many copies of $\mathbb{Z}_k$
$$\mathbb{Z}_k\wr S_{c_k(\sigma)}=(\underbrace{\mathbb{Z}_k\times\cdots\times\mathbb{Z}_k}_{c_k(\sigma)})\rtimes S_k$$
are generated by the $k$-cycles of $\sigma$, and the elements of $S_{c_k(\sigma)}$ "block"-permute the cycles themselves.
A low-tech way to understand this is to write cycles of common length (in cycle notation) in rows to form rectangles. For instance, if $\sigma=(123)(456)$ then we may write the rectangle
$$ \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array} $$
The elements of $C(\sigma)\subset S_6$ are generated by those that cycle the labels of individual rows and permutations of the rows themselves (preserving the order of labels within the rows). That is, $C(\sigma)$ is generated by $\mathbb{Z}_3\times\mathbb{Z}_3=\langle(123),(456)\rangle$ and $S_2=\{e,(14)(25)(36)\}$.
Exercise. $C(\sigma)\subseteq A_n$ if and only if $\sigma$ is a product of distinct odd-length cycles.