If $G'$ has order less than $p^3$, then it is abelian. So we may assume that $G'$ has order exactly $p^3$, in which case $G/G'$ is of order $p^2$, and thus $2$-generated. let $x$ and $y$ be the two generators. Moreover, $G_3=[G,G']$ has order at most $p^2$, $G_4$ has order at most $p$, and so $G_5$ will certainly be trivial; that is, $G$ has class at most $4$.
Since $G$ is a finite $p$-group, $G'$ is generated by the basic commutators in $x$ and $y$. We know that $G_n/G_{n+1}$ is generated by the basic commutators of weight $n$. So $G_2/G_3$ is cyclic, generated by $[y,x]$; $G_3/G_4$ is generated by $[y,x,x]$ and $[y,x,y]$. And $G_4/G_5 = G_4$ is generated by the basic commutators of weight $4$; but the only basic commutators of weight $4$ that are nontrivial are $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$, since there is no nontrivial basic commutator obtained as $[c,c']$ with $c$ and $c'$ of weight $2$.
Thus, $G'$ is generated by $[y,x]$, $[y,x,x]$, $[y,x,y]$, $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$. Since $G'$ is of class at most $4$, they commute pairwise (the only commutator not of weight greater than $4$ is $[[y,x],[y,x]]$, which is trivial).
Suppose $N \leq G$ is normal, and $A$ is not contained in $N$, nor is $B$. By normality of $N$ and $A$, $[N,A] \leq A \cap N$, and $A \cap N$ is normal in $A$. By simplicity, $[N,A]=1,$ so for all $n \in N, a \in A,$ $nan^{-1}a^{-1}=1,$ and therefore $(n^{-1})^{a}=n^{-1}$, which implies that $A \subset C_G(N)$. Similarly, $B \subset C_G(N)$. It follows that $N \leq Z(G).$
Assuming that $A,B$ are nonabelian, we get that $Z(G)=Z(A)\times Z(B)=1.$ Therefore, $N$=1. The case of $A \lneq N$ should contradict the simplicity of $B$.
Best Answer
You are quite correct. It is first of all true that for any set $A$ the derived subgroup of the finitary symmetric group on $A$ (the group of all permutations of finite support) has the alternate group on $A$ as its derived subgroup.
In the particular case of $\mathrm{A}_4$, since you know that the Klein-type subgroup $K$ is normal and that any element of $\mathrm{A}_4$ is either a bi-transposition (term coined up ad-hoc by me by which I mean a product of $2$ disjoint transpositions) in $K$ or a $3$-cycle, you only need to assess the commutator between two $3$-cycles (a commutator which has at least one term in $K$ will itself belong to $K$ by normality); two $3$-cycles of equal support are either equal or one is the square of the other, so at any rate they commute and their commutator is therefore trivial; if not, then by the inclusion-exclusion principle their supports intersect necessarily in a two element set $\{a, b\}$, such that one of the cycles is of support $\{a, b, c\}$ and the other one of support $\{a, b, d\}$; without any loss of generality then the commutator studied will be either of the type: $$[(abc)(abd)]=(ab)(cd)$$ or $$[(abc)(bad)]=(ac)(bd)$$ and at any rate will belong to $K$. Hence, on the one hand $\mathrm{D}(\mathrm{A}_4) \leqslant K$ and the reverse inclusion can be established either by your line of reasoning or by noticing that the above commutation relations express all elements of $K$ as commutators between some pairs of $3$-cycles.