Let $C$ be a category with a terminal object $\ast$. For two pointed objects, define their wedge sum $c\vee d$ as the pushout $$ \require{AMScd} \begin{CD} \ast @>>c_0> c \\ @VVd_0V@VV\iota_cV \\ d @>>\iota_d> c \vee d \end{CD} $$ $\iota_c$, $\iota_d$ are sections.
Using the universal properties of $c \vee d$ and $c \times d$, we can now define a canonical map $c \vee d\overset{n}\to c \times d$ such that $$\pi_cn\iota_c = 1_c,\ \pi_dn\iota_d = 1_d,$$ $$\pi_dn\iota_c = d_0\ast_c,\ \pi_cn\iota_d = c_0\ast_d,$$ (where I'm using $\ast_x$ as a shorthand for the morphism $x\to\ast$).
Is $n:c \vee d\to c \times d$ a section? Is $n$ monic?
The answer is yes when $C=Set$, and I suspect it's very basic, or false, in general...
Not in general.
In $\mathsf{Group}$, every element is pointed, with zero object the trivial group. The wedge is just the free product $G*H$, and the product is the direct product $G\times H$.
The map $n$ you describe is the map $n\colon G*H\to G\times H$ obtained by moding out by the cartesian subgroup, $$[G,H] = \langle [g,h]\mid g\in G,h\in H\rangle.$$ If $G$ and $H$ are nontrivial, then the map has nontrivial kernel (since no nontrivial element of $G$ commutes with a nontrivial element of $H$ in $G*H$). Thus, $n$ cannot be monic (in this category, monic is equivalent to one-to-one), and hence cannot have a right inverse, so it is not a section either.
On the other hand, in an abelian category you get an isomorphism.