Existence of homotopy inverses for co-H spaces Suppose $c: X \to X\vee X$ is a co-$H$ structure on a based CW complex $X$. 
 Question: Under what circumstances can one find left (right) homotopy inverses for $c$?
 Remarks:  If $X$ is $1$-connected, then the answer is yes (for a proof, see below).
But I don't know the answer in general. A good test case would be to take an exotic comultiplication on the circle and see what happens there.
A reduced suspension of a based space, with its tautological comultiplication (the pinch map), has homotopy inverses. These arise from the reflection map on the circle coordinate. 
 Definitions:   a  co-$H$ structure on $X$ is a map $c$ as above such that
the composite
$$
X \quad \overset{c}\to \quad X \vee X \quad \overset{\text{include}}\longrightarrow \quad X \times X
$$ 
is homotopic to the diagonal.  A  left homotopy inverse  for $c$ is a map
$\ell: X \to X$ such that the composite
$$
X \quad \overset{c} \to \quad X\vee X\quad  \overset{\ell \vee \text{id}}\longrightarrow \quad X\vee X
\quad \overset{\text{fold}}\longrightarrow \quad X
$$
is homotopic to a constant map. (Right homotopy inverses are defined similarly.)
  proof when $X$ is $1$-connected:  Form the  coshearing map 
$$
\check S : X \vee X \to X \vee X
$$
which on the first summand of the domain is given by $c$ and on the second by the second summand inclusion.  It is straightforward to check that $\check S$ is a homology isomorphism,
so by the Whitehead theorem, there's a map $T: X\vee X \to X\vee X$ such that 
$\check S\circ T$ and $T\circ \check S$ are each homotopic to the identity map.
Then the composite 
$$
X \quad \overset{i_1}\to \quad X \vee X \quad \overset{T}\to \quad X\vee X \quad \overset{p_2}\to \quad X
$$
is a left homotopy inverse for $c$, where $i_1$ is the first summand inclusion and $p_2$ is the projection onto the second summand.
 A: The argument you give for the simply-connected case generalizes to nilpotent connected co-H spaces. This is essentially because nilpotent spaces are $H_\ast(-;\mathbb{Z})$-local. See
Hilton, Peter; Mislin, Guido; Roitberg, Joseph On co-H-spaces. Comment. Math. Helv. 53 (1978), no. 1, 1–14,
in particular Theorem 2.3 and the remarks just after. However, in the same paper they show that the only non-simply-connected nilpotent connected co-H-space is $S^1$, and remark that "it is easy to see that any co-H-structure on $S^1$ admits a 2-sided co-inverse".
I think that if $X$ and $Y$ are co-H-spaces which admit left (right) co-inverses $\ell_X\colon\thinspace X\to X$ and $\ell_Y\colon\thinspace Y\to Y$, then $X\vee Y$ with the co-H-structure
$$
X\vee Y \stackrel{c_X\vee c_Y}{\longrightarrow} (X\vee X)\vee(Y\vee Y)\cong (X\vee Y)\vee(X\vee Y)
$$
has as a left (right) co-inverse the map $\ell_X\vee\ell_Y$. However this does not seem to be enough to answer your question about a finite wedge of circles, since a given co-H-structure may not split as a wedge in this way.
