As you probably already know, a monoid object $M$ in the category of sets with Cartesian product is precisely a monoid in the usual sense: the unit map $u: * \to M$ takes the single-element set to the unit element of the monoid, while the multiplication $m \colon M \times M \to M$ gives the monoid multiplication.
What is a comonoid object $C$ in this category? There is only a single possibility for the counit map $\epsilon \colon C \to *$, so we need to only work out the comultiplication $\Delta \colon C \to C \times C.$ Since $\times$ is the categorical product, $\Delta$ is equivalent to two maps $l: C \to C$ and $r: C \to C$ such that $\Delta(c) = (l(c), r(c))$ for all $c \in C$.
The left counit axiom says that $c \mapsto (*, c)$ should be equivalent to $c \mapsto (l(c), r(c)) \mapsto (*, r(c))$. Hence we find that $r: C \to C$ must be the identity function. Similarly for $l$, and so $\Delta \colon C \to C \times C$ must simply be the map $\Delta(c) = (c, c)$. It is easy to see that $\Delta$ is coassociative.
Checking the bimonoid axioms, every monoid object $(M, u, m)$ automatically becomes a bimonoid object $(M, u, m, \epsilon, \Delta)$ with $\epsilon$ and $\Delta$ as above. So all that is left is to figure out what antipode map $S \colon M \to M$ needs to satisfy. One of the Hopf axioms is that the composition
$$ x \mapsto (x, x) \mapsto (x, S(x)) \mapsto m(x, S(x))$$
is equal to the composition
$$ x \mapsto * \mapsto 1$$
where $1$ is the unit in the monoid. So $S(x)$ must be a right inverse for $x$. The other Hopf axiom gives that $S(x)$ is a left inverse for $x$.
So indeed, a monoid object in the category of sets is always a bimonoid object, and this can be equipped with an antipode if and only if every element of the monoid is invertible.