To begin with, my personal opinion and experience is that the name ''group completion theorem'' is quite bombastic and mainly a source of confusion. This is not to say that one cannot \emph{interpret} the results as ''group completions'' in a useful way.
I understood the story when I read the last chapter of Galatius, Madsen, Tillmann, Weiss ''The homotopy type of the cobordism category''. It is a superb exposition, not least because the geometric and algebraic-topological arguments are cleanly separated.

In all the results you quote, the ''scanning'' (or Pontrjagin-Thom, or h-principle)arguments and the ''group completion'' arguments belong to really different parts of the story.

To explain these two different parts, let us go through the classical example of configuration spaces of points in $\mathbb{R}^n$.

The ''scanning'' part of the argument: Let $C^{n,k}$ be the space of configurations in $R^n$, contained in $I^k \times R^{n-k}$, with the topology that Hatcher describes. There are several maps that could be called scanning maps in this context. One is $C^{n,k} \to \Omega C^{n,k-1}$ and it is a homotopy equivalence if $k \leq n-1$. Another one is the homotopy equivalence $C^{n,0} \to S^n$. Both together give homotopy equivalences $C^{n,k} \to \Omega^k S^n$; both can be derived also using Gromov's general h-principle. There are no homology, no monoids or categories, let alone ''group completion'' in that part.

''Group completion'' comes in when you consider the scanning map $ C^{n,n} \to \Omega C^{n,n-1}$; it is not a homotopy equivalence. It can be stabilized and the stabilized map is a map $Z \times Conf^{\infty}(R^n) \to \Omega C^{n,n-1}$, which is a homology isomorphism.

But you asked for other examples; a classical one is Quillen's algebraic $K$-theory; and there is no ''scanning map'', in absence of a geometric model. Let $R$ be a ring and let $Proj(R)$ be a set of representatives for the isomorphism classes of finitely generated projective $R$-modules and $M:= \coprod_{P \in Proj(R)} B Aut(P)$. The statement is that there is a homology isomorphism

$$K_0 (R)\times B GL_{\infty}(R) \to \Omega BM.$$

This is the analogue of the statement on $Z \times Conf^{\infty}(R^n) \to \Omega C^{n,n-1}$, but here there is no way to interpret it as a scanning map. The statement $ \Omega C^{n,n-1} \simeq \Omega^n S^n$ does not have an analogue; basically because there is no real geometric model for $B GL_n (R)$ (the $K$-theory spectrum is no Thom spectrum!).

The proof begins with turning $M$ into something that you can take $B$ of, say a monoid under direct sum. Without a geometric model, this is not a complete triviality. Then there is a self-map $M \to M$, given by adding a trivial rank one $R$-module and form
$M_{\infty} = colim (M \to M \to \ldots ) \simeq K_0 (R)\times B GL_{\infty}(R) $.

$M$ ''acts'' on $M_{\infty}$ and you can form $EM \times_M M_{\infty}$ (after you made sense out of this). There is a map $f:EM \times_M M_{\infty} \to BM$. The source of $f$ is contractible, and hence the homotopy fibre of $f$ is $\Omega BM$. The actual fibre of $f$ is $M_{\infty}$ and these observations yield a map $M_{\infty} \to \Omega BM$, which we claim is a homology isomorphism.

The translation maps $M_{\infty} \to M_{\infty}$ given by multiplication by elements of $M$ are homology equivalences. This step is easy in the situation of modules; but for the Madsen-Weiss or the Galatius theorem, the analogous argument requires homological stablity. Because the translations are homology equivalences, $f$ is a homology fibration. By MacDuff-Segal, the natural map from the actual fibre of $f$ to the homotopy fibre is a homology isomorphism, which completes the proof.
I like to think about this last step as a local-to-global principle, similar in spirit to, say, the result that a fibre bundle (local condition) is a fibration (global condition).