No, you cannot even do it in groups. For if you could, consider the identity $x^n=e$ with $n>1$. If this were equivalent to a set of commutator words, then every abelian group would satisfy the relation, but of course this is false: only those abelian groups that are of exponent $n$ satisfy it.
More generally, in groups every identity is equivalent to a pair of identities, one a commutator word and one a power word; equivalently, if $w$ is a word in the free group $F$ on countably many generators $x_1,\ldots,x_n,\ldots$, then there exist words $w_1$ and $w_2$, with $w_1=x_1^a$ for some positive integer $a$, and $w_2\in [F,F]$, such that for all groups $G$, $G$ satisfies $w$ as an identity (every evaluation of $w$ in $G$ is trivial) if and only if $G$ satisfies $w_1$ and $w_2$ as identities. This is due to Bernhard Neumann; write $w=x_1^{n_1}\cdots x_m^{n_m}c$, with $c$ a commutator word, by using the collection process. Then let $a=\gcd(n_1,\ldots,n_m)$, and $w_2=c$.
So at least in groups you also need "power identities".