I have asked myself almost exactly the same question but haven't found any answer in the literature. Then I have found a proof which ended up being very similar to the classical one. This might be the reason why this approach is not discussed anywhere, perhaps people who found it just didn't feel that it contributed any more insight than the standard proof. Besides that, I can think of only one (but important) advantage of the normalized chain complex: it makes it easy to prove that $\pi_* A \cong H_* N A$.

Here's an argument along the lines you suggest.

By the triangular identities it is enough to check that the unit $\eta_A \colon A \to \Gamma C A$ is an isomorphism and that $\Gamma$ reflects isomorphisms.

Let's do the latter first. For $m > 0$ let $D^m$ denote the chain complex freely generated by $\iota$ in degree $m$ and by the boundary of $\iota$ in degree $m - 1$ and zero elsewhere. Then chain maps out of $D^m$ classify $m$-chains. The map $D^m \to C\mathbb{Z}\Delta[m]$ classifying the identity has a retraction given by sending $\mathrm{id}_{[m]}$ to $\iota$, $\delta_0$ to the boundary of $\iota$ and everything else to zero. Complexes $D^m$ along with $\mathbb{Z}$ concentrated in degree $0$ detect isomorphisms and hence so do all $C\mathbb{Z}\Delta[m]$s.

Now consider the unit $\eta_A \colon A \to \mathrm{Ch}(C\mathbb{Z}\Delta[m], CA)$. It acts by regarding $a \in A_m$ as $a \colon \mathbb{Z}\Delta[m] \to A$ and applying $C$. I will construct the inverse map sending $f \colon C\mathbb{Z}\Delta[m] \to CA$ to $\tilde{f} \in A_m$ by induction on $m$. The main ingredient is the same as in the standard proof, i.e. a lemma saying that in $A_m$ the subgroups $NA_m \bigcap_{i < m} \ker \delta_i$ and $DA_m = \Sigma_{i < m} \mathrm{im}\, \sigma_i$ intersect trivially. Start with $f \colon C\mathbb{Z}\Delta[0] \to CA$ and set $\tilde{f} = f(\mathrm{id}_{[0]}) \in CA_0 = A_0$. Next, assume that we have handled all simplices below degree $m$ and let $f \colon C\mathbb{Z}\Delta[m] \to CA$. Then $f(\mathrm{id}_{[m]}) \in CA_m = A_m / DA_m$, pick $a \in A_m$ such that $f(\mathrm{id}_{[m]}) = a + DA_m$. The simplex $\tilde{f}$ that we want to construct has to live in the same coset, i.e. $\tilde{f} = a + x$ for some $x \in DA_m$. Moreover, we need $\tilde{f} \delta_i = \widetilde{f \delta_i}$ for all $i \in [m]$ where $\widetilde{f \delta_i}$s were constructed in the previous step of the induction. The lemma I mentioned before says that there is a unique $x$ such that $\tilde{f} \delta_i = \widetilde{f \delta_i}$ for all $i < m$. We need to check that we also have $\tilde{f} \delta_m = \widetilde{f \delta_m}$ and this follows by induction since both $\tilde{f} \delta_m$ and $\widetilde{f \delta_m}$ are determined by their boundaries and reductions modulo $DA_{m-1}$. Thus $\tilde{f}$ is a unique simplex with $\eta_A(\tilde{f}) = f$.

I should also mention that it is possible to phrase the standard argument in a similar way, except that it is easier to exhibit $N$ as a right rather than a left adjoint of $\Gamma'$. (I'm going to write $\Gamma'$ to distinguish it from $\Gamma$ above.) The unit $\eta'_X \colon X \to N \Gamma' X$ is given by $\eta'_X(x) = (\mathrm{id}, x)$ and the counit $\epsilon'_A \colon \Gamma' N A \to A$ is given by $\epsilon'_A(\sigma, a) = a \sigma$. Then the triangular identities are immediately verified and the standard proof can be arranged to say that $\eta'_A$ is an isomorphism and $N$ reflects isomorphisms.