A well-written discussion of the group completion can be found on pp. 89--95 of
J.F. Adam: Infinite loop spaces, Ann. of Math. studies 90 (even though he only
discusses a particular group completion of a monoid). In particular you
assumption of commutativity comes in under the assumption that $\pi_0(M)$ is
commutative which makes localisation with respect to it well-behaved
(commutativity is not the most general condition what is needed is some kind of
Øre condition).
In any case if you really want conclusions on the homotopy equivalence level I
think you need to put yourself in some nice situation for instance requiring
that all spaces be homotopy equivalent to CW-spaces. If you don't want that you
should replace homotopy equivalences by weak equivalences, if not you will
probably find yourself in a lot of trouble. In any case I will assume that we
are dealing with spaces homotopy equivalent to CW-complexes.
Starting with 1) a first note is that your conditions does not have to involve
an arbitrary ring $R$. It is enough to have $R=\mathbb Z$ and one should
interpret the localisation in the way (for instance) Adams does:
$H_\ast(X,\mathbb Z)=\bigoplus_\alpha H_\ast(X_\alpha,\mathbb Z)$, where
$\alpha$ runs over $\pi_0(X)$, and a $\beta$ maps $H_\ast(X_\alpha,\mathbb Z)$
to $H_\ast(X_{\alpha\beta},\mathbb Z)$. Then your group completion condition is
that the natural map $\mathbb Z[\pi_0(Y)]\bigotimes_{\mathbb Z[\pi_0(X)]}
H_\ast(X,\mathbb Z)\rightarrow H_\ast(Y,\mathbb Z)$ should be an isomorphism.
This then implies the same for any coefficient group (and when the coefficient
group is a ring $R$ you get your condition). (Note that for this formula to even make sense we need at least associativity for the action of $\pi_0(X)$ on the homology. This is implied by the associativity of the Pontryagin product of $H_*(X,\mathbb Z)$ which in turn is implied by the homotopy associativity of the H-space structure.)
Turning now to 1) it follows from standard obstruction theory. In fact maps into
simple (hope I got this terminology right!) homotopy types, i.e., spaces for
which the action of the fundamental groups on the homotopy groups is trivial (in
particular the fundamental group itself is commutative). The reason is that the
Postnikov tower of such a space consists of principal fibrations and the lifting
problem for maps into principal fibrations is controlled by cohomology groups
with ordinary coefficients. Hence no local systems are needed (they would be if
non-simple spaces were involved). The point now is that H-spaces are simple so
we get a homotopy equivalence between any two group completions and as
everything behaves well with respect to products these equivalences are H-maps.
Addendum:
As for 2) it seems to me that this question for homotopy limits can only be solved under supplementary conditions. The reason is that under some conditions we have the Bousfield-Kan spectral sequence (see Bousfield, Kan: Homotopy limits, completions and localizations, SLN 304) which shows that $\varprojlim^s(\pi_s X_i)$ for all $s$ will in general contribute to $\pi_0$ of the homotopy limit. As the higher homotopy groups can change rather drastically on group completion it seems difficult to say anything in general (the restriction to cosimplicial spaces which the OP makes in comments doesn't help as all homotopy limits can be given as homotopy limits over $\Delta$. Incidentally, for homotopy colimits you should be in better
shape. There is however an initial problem (which also exists in the homotopy limit case): If you do not assume that the
particular group completions you choose have any functorial properties it is not
clear that a diagram over a category will give you a diagram when you group
complete. This can be solved by either assuming that in your particular
situation you have enough functoriality to get that (which seems to be the case
for for instance May's setup) or accepting "homotopy everything" commutative
diagrams which you should get by the obstruction theory above. If this problem
is somehow solved you should be able to conclude by the Bousfield-Kan spectral
sequence $\injlim^\ast H_*(X_i,\mathbb Z)\implies
H_*(\mathrm{hocolim}X_i,\mathbb Z)$. We have that localisation is exact and
commutes with the higher derived colimits so that we get upon localisation a
spectral sequence that maps to the Bousfield-Kan spectral sequence for $\{Y_i\}$
and is an isomorphism on the $E_2$-term and hence is so also at the convergent.
As for 3) I don't altogether understand it. Possibly the following gives some
kind of answer. For the H-space $\coprod_n\mathrm{B}\Sigma_n$ which is the
disjoing union of classifying spaces of the symmetric groups its group
completion has homotopy groups equal to the stable homotopy groups of spheres
which shows that quite dramatic things can happen to the homotopy groups upon
group completion (all homotopy groups from degree $2$ on of the original space
are trivial).
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.
Best Answer
The statement that $M \to \Omega BM$ is a weak equivalence when $M$ is a group-like topological monoid is indeed easier: the map $EM = B(M \wr M) \to BM$ is then a quasi-fibration, has geometric fibre $M$ over the basepoint and homotopy fibre $\Omega BM$.
However the homological group-completion theorem also implies this: if $M$ is group-like then $\pi_0(M)$ already consists of units in $H_*(M)$, so it just says that $M \to \Omega BM$ is a homology equivalence. Each of these spaces has homotopy equivalent path components, so it is then enough to observe that the map of 0 components is a homology equivalence between simple spaces, so a weak homotopy equivalence.
However it is perverse to prove the "$M \simeq \Omega BM$" result this way.