I think I have been able to reproduce the "argument by Wagoner" (perhaps it was removed from the published version?). It certainly holds in more generality that what I have written below, using the notion of "direct sum group" in Wagoner's paper (which unfortunately seems to be a little mangled).
Let $M$ be a homotopy commutative topological monoid with $\pi_0(M)=\mathbb{N}$. Choose a point $1 \in M$ in the correct component and let $n \in M$ be the $n$-fold product of 1 with itself, and define $G_n = \pi_1(M,n)$. The monoid structure defines homomorphisms $$\mu_{n,m} : G_n \times G_m \longrightarrow G_{n+m}$$ which satisfy the obvious associativity condition. Let $\tau : G_n \times G_m \to G_m \times G_n$ be the flip, and $$\mu_{m,n} \circ \tau : G_n \times G_m \longrightarrow G_{n+m}$$ be the opposite multiplication. Homotopy commutativity of the monoid $M$ not not ensure that these two multiplications are equal, but it ensures that there exists an element $c_{n,m} \in G_{n+m}$ such that $$c_{n,m}^{-1} \cdot \mu_{n,m}(-) \cdot c_{n,m} = \mu_{m,n} \circ \tau(-).$$
Let $G_\infty$ be the direct limit of the system $\cdots \to G_n \overset{\mu_{n,1}(-,e)}\to G_{n+1} \overset{\mu_{n+1,1}(-,e)}\to G_{n+2} \to \cdots$.
Theorem: the derived subgroup of $G_\infty$ is perfect.##
Proof: Let $a, b \in G_n$ and consider $[a,b] \in G'_\infty$. Let me write $a \otimes b$ for $\mu_{n,m}(a, b)$ when $a \in G_n$ and $b \in G_m$, for ease of notation, and $e_n$ for the unit of $G_n$.
In the direct limit we identify $a$ with $a \otimes e_n$ and $b$ with $b \otimes e_n$, and we have $$b \otimes e_n = c_{n,n}^{-1} (e_n \otimes b) c_{n,n}$$ so $b \otimes e_n = [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)$. Thus $$[a \otimes e_n, b \otimes e_n] = [a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)]$$ and because $e_n \otimes b$ commutes with $a \otimes e_n$ this simplifies to $$[a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)]].$$ We now identify this with $$[a \otimes e_{3n}, [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}]$$ and note that $a \otimes e_{3n} = c_{2n,2n}^{-1}(e_{2n} \otimes a \otimes e_{n})c_{2n,2n} = [c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})]\cdot (e_{2n} \otimes a \otimes e_{n})$. Again, as $(e_{2n} \otimes a \otimes e_{n})$ commutes with $[c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}$ the whole thing becomes $$[a,b]=[[c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})], [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}],$$ a commutator of commutators.
