Suppose that it's unknown whether a category $\mathcal{C}$ contains all products, but that it happens to have the products $a\times b$ and $c\times d$. Further, $\mathcal{C}$ has morphisms $f:a\rightarrow c$ and $g:b\rightarrow d$. In what situations must the product map $f\times g$ exist? (Or something isomorphic to it, since we aren't necessarily dealing with pointwise functions.)
Since $\{f\circ \pi_a:a\times b\rightarrow c,g\circ \pi_b:a\times b\rightarrow d\}$ is a cone over $\{c,d\}$, some $h:a\times b\rightarrow c\times d$ must exist as a factorization through $c\times d$, i.e., $f\circ \pi_a=\pi_c\circ h$ and $g\circ \pi_b=\pi_d\circ h$. But does this mean that $h\cong f\times g$?
As a special case, suppose that $\mathcal{C}$ has a terminal object $1$, the products $1\times 1$ (not necessarily terminal) and $a\times b$, and element selectors $x:1\rightarrowtail a$ and $y:1\rightarrowtail b$. When must $x\times y$ exist (up to isomorphism)? Stated colloquially: In what situations can you construct an element selector for a product by combining element selectors for its constituent objects?
Motivation: My intuition for set products has long been that you can construct an element by selecting one element from each constituent set. I've recently realized, however, that elements of non-set products might not be constructible from smaller parts. The categorical definition only requires that they can be projected onto their constituent objects.
As the comments lay out, $f\times g$ exists and is given by your $h$ once $a\times b$ and $c\times d$ exist. To see this, it's important to first understand what $f\times g$ means.
Given $f:a\to c$ and $g:b\to d$, the product $f\times g$ should map $a\times b\to c\times d$ in a way that resembles "acting by $f$ on $a$ and acting by $g$ on $b$ and then combining them." In categorical language, this means exactly that $\pi_c(f\times g)=f\pi_a$ and $\pi_d(f\times g)=g\pi_b$.
Let's return to your $h$. You've shown that once $a\times b$ and $c\times d$ exist, then there is a map $h:a\times b\to c\times d$ with the same property that $\pi_ch=f\pi_a$ and $\pi_dh=g\pi_b$. The key here is that since $h$ was defined by the universal property of $c\times d$, it is unique with this property, and this uniqueness forces that $f\times g=h$. In other words, your construction describes precisely how to obtain $f\times g$ from $f$ and $g$.
In your special case where $a=b=1$, this shows that the answer is always! In fact, since $1\times1=1$ (check universal properties), the resulting map $x\times y:1\times 1\to c\times d$ corresponds exactly to the tuple $(x,y):1\to c\times d$ obtained from the universal property of $c\times d$.
To address your motivation: you actually can construct products in an arbitrary category by combining its constituent "elements" once you establish what "elements" means. From the perspective of the Yoneda embedding, you can equivalently think of an object $a$ of a (locally small) category as a family of sets $(a_s)_s$ indexed by the category, where $a_s$ is the set of "$s$-shaped elements in $a$"; that is, morphisms $s\to a$. (These elements are then subject to compatibility constraints which just say that $s\mapsto a_s$ is a presheaf, since this is really the representable functor $\def\Hom{\operatorname{Hom}}\Hom(-,a)$).
What is a product from this perspective? Given two objects $a$ and $b$, the product $a\times b$ should be the collection of pairs of elements of $a$ and $b$: in other words, we would want $(a\times b)_s = a_s\times b_s$, the right hand side being the usual Cartesian product of sets. If the product $a\times b$ exists, then this is indeed true by the universal property of the product! (In fact, this could be seen as a way of motivating the universal property definition.) Indeed, it says that for any $s$-shaped elements $x:s\to a$ and $y:s\to b$, there must be a unique corresponding element $(x,y):s\to a\times b$.
This is also a way of making sense of the statement that "$a\times b$ is, when it exists, the representing object of the presheaf $\Hom(-,a)\times\Hom(-,b)$."