A topological space $X$ which is an algebra over an $E_\infty$-operad $E$ consists of a sequence of maps $$ \mu_n':E(n)\times X^n\to X $$ with compatibility conditions. The spaces $E(n)$ are contractible and the symmetric group $\Sigma_n$ acts freely on them.
If one would have a strictly associative operation $\mu:X\times X\to X$ then by iterating, one would get a map $$ \mu_n:X^n\to X $$ which factors, if $\mu$ is strictly commutative through $X^n/\Sigma_n$. Here, the group $\Sigma_n$ acts on $X^n$ by commuting the factors.
I guess, as suggested in the accepted answer to this MO question, that for an algebra $X$ over an $E_\infty$-operad $E$ (hence something ''weaker'' than a space with a strictly commutative operation), one gets a factorization of $\mu_n'$ through $(E(n)\times X^n)/\Sigma_n$.
Why does one get such a factorization of $\mu_n'$? Is it ''obvious'' from the definition of an algebra over an operad?
It seems I confused two different kinds of $\Sigma_n$-invariance in my (now deleted) comment. In fact the factorisation you want is indeed just the compatibility condition for algebras over the symmetric operad. Indeed, given any symmetric operad $E$, the definition of $E$-algebra includes the condition $$\sigma (e) \cdot (x_{\sigma (1)}, \ldots, x_{\sigma (n)}) = e \cdot (x_1, \ldots, x_n)$$ for all $\sigma$ in $\Sigma_n$. (The $\Sigma_n$-actions on $E (n)$ and $X^n$ are both right actions, despite the notation.) This is precisely the statement that the structure map $E (n) \times X^n \to X$ is $\Sigma_n$-invariant, which implies that it factors through the coinvariants.