Try approaching this via induction on $n$?
You know that $(1,2,3)$ generates $A_3$ and $(1, 2, 3)(3,4,5)$ is a $5$-cycle, so you can generate $A_5$ on $\{1,2,3,4,5\}$. So you have $3$-cycles of the form $(a,b,5)$ with $a,b\in\{1,2,3,4\}$, and that lets you generate $A_5$. Then note that in $\{a,b,5,6,7\}$ you can find the needed 3-cycles on $\{1,2,3,4,5,6,7\}$ that generate $A_7$...
Assume this holds in terms of the three-cycles generating $A_{2n - 1}$, then show it holds for $A_{2(n+1)-1} = A_{2n+1}.$
Let $r\in[1\>..\>n-1]$ be the length of the cycle containing the number $n$. Begin the listing of this cycle with $n$. There are $(n-1)(n-2)\cdots(n-r+1)$ ways to choose the remaining entries of this cycle. Now there will be $n-r\geq1$ numbers left over, one of them the largest. Begin the listing of the second cycle with this largest left-over number, and you will then have $(n-r-1)!$ ways to choose the remaining entries of the second cycle. In all we have ${(n-1)!\over n-r}$ permutations of this kind. It follows that there are
$$a_n=(n-1)!\sum_{k=1}^{n-1}{1\over k}\tag{1}$$
elements of $S_n$ having exactly two cycles. In particular this gives $a_4=11$. – Maybe the formula $(1)$ can be brought into a closed form, using some combinatorial identity.
Computing the $a_n$ we obtain the sequence $(1, 3, 11, 50, 274, 1764, 13068, 109584, 1026576,\ldots)$ which OEIS identifies as A000254.
Best Answer
A permutation is an element of $A_n$ if and only if it is a product of an even number of transpositions.
We first note that the three cycles do not generate more than $A_n$ since for distinct $i,j,$ and $k$, we have $(ij)(ik)=(ijk)$.
For the other inclusion, we note that $(ijk)=(ij)(ik)$ for distinct $i,j,$ and $k$. And for distinct $i,j,k,$ and $l$, we have that $(ij)(kl)=(ijk)(jkl)$. Thus the three cycles generate all products of an even number of transpositions.