I feel like the key observation here is that in some categories (like groups) we have a canonical map from coproducts to products, induced by these inclusion maps that you've noticed exist. The natural question is where does it come from? Perhaps answering this question will shed some light on what's going on here, since we can then take a look at other examples of this phenomenon to get intuition.
Motivation: (This section is sorta fuzzy mathematically in places to motivate the definitions without worrying about details)
So what is a map from the coproduct to the product? Well by definition, if $\newcommand\C{\mathcal{C}}\C$ is our category, then
$$
\newcommand\of[1]{\left({#1}\right)}
\C\of{
\coprod_i X_i,
\prod_j X_j
}
\cong
\prod_{i,j}\C(X_i,X_j).
$$
So a map from the coproduct to the product requires for every pair of objects
a choice of map $\C(X,Y)$. When $X=Y$, this is easy, we can take the identity map.
What do we do if $X\ne Y$ though? Well, if we require our category has finite products/coproducts including initial and terminal objects, $0$ and $1$ respectively, then if the unique map $0\to 1$ is an isomorphism, then we can always produce a map
$$X\to 1\to 0\to Y,$$
and this doesn't depend on our choice of $0$ or $1$, since everything is unique up to isomorphism.
This gives us a definition.
Categories with zero objects:
If $0\to 1$ is an isomorphism, then $0$ is both initial and terminal, and we say $\C$ has a zero object, and from now on I'll write $0$ for a zero object.
We also say that the unique map $X\to 0\to Y$ is the zero morphism from $X$ to $Y$, written as $0$. (This gives a canonical enriching in pointed sets with smash product, which is related to your observation that groups have a distinguished element).
Thus in a category with zero objects, we can define a canonical morphism
$$
\coprod_i X_i \to \prod_j X_j,
$$
with components $1_{X_i}$ when $i=j$ and $0$ when $i\ne j$. (Writing this as a matrix, you'll note that this is the identity matrix).
However, this canonical map is not generally an isomorphism. When it is (for finite sums/products), we call the object the biproduct, written $X\oplus Y$, and in such a situation, we get a canonical enriching in commutative monoids. The addition of $f,g : X\to Y$ is given by the composite
$$ X\xrightarrow{\Delta} X\oplus X \newcommand\toby\xrightarrow\toby{f\oplus g}
Y\oplus Y \toby{\nabla} Y. $$
Examples:
The trivial group, $1$, is the zero object in both groups and abelian groups, and you can check that the canonical morphism I define above gives you the same result as the morphism induced by the universal property given by your inclusions. I.e., it sends
$g \in G_i$ to the tuple $(1,1,\ldots,1,g,1,\ldots,1)$, with $g$ in the $i$th spot.
For another category, you have the category of pointed sets or pointed topological spaces (pairs $(X,x)$ with $x\in X$ and morphisms $f:(X,x)\to (Y,y)$ are the maps $f:X\to Y$ such that $f(x)=y$).
The coproduct here is called the wedge sum, and it naturally includes into the product in the same way, with $x\in X_i$ mapping to
$(*,\ldots,*,x,*,\ldots,*)$, where $x$ is in the $i$th location, and
$*$ is the basepoint in the other factors.
Finally, let's give a bit of a weird one. (Though this is a category I came across recently.)
$R$-algebras (for me right now, the category of commutative, unital algebras over a commutative ring $R$) do not have zero objects (unless $R=0$). The initial object is $R$, and the terminal object is the zero ring. However, we can consider the category of
$R$-algebras with augmentations. Explicitly, these are commutative rings $S$ with
maps
$$ R\toby{\iota_S} S \toby{\pi_S} R,$$
where $\pi_S\iota_S = \newcommand\id{\mathrm{id}}\id_R$.
Morphisms are ring maps $\phi : S\to T$ such that $\pi_T\phi = \pi_S$ and
$\phi\iota_S = \iota_T$. Now $R$ is a zero object in this category. (This is a special case of a general way to produce new categories with zero objects, make the objects either pairs $(X,1\to X)$ of an object and a morphism from the terminal object or the dual construction, take pairs $(X,X\to 0)$, which is what we did here). The first construction is common, and called taking the pointed category of $\C$, denoted $\C_*$, and is what we do to produce pointed sets and pointed topological spaces.
The coproduct of $S$ and $T$ is $S\otimes_R T$, with augmentation given by $(s\otimes t)\mapsto \pi_S(s)\pi_T(t)$. The product of $S$ and $T$ is given by (the fiber product) $S\times_R T$, with algebra structure map given by $r\mapsto (\iota_S(r),\iota_T(r))$.
Then the morphism $S\otimes_R T\to S\times_R T$ is given by $s\mapsto (s,\iota_T(\pi_S(s))$ and $t\mapsto (\iota_S(\pi_T(t)),t)$.
Conclusion
Hopefully I've given a bit of context and background to the construction that you've noticed. I hope you'll see that asking the question of why the product fails to be the coproduct in groups, while a reasonable question, doesn't have much of an answer other than: because it can't be.
While that can feel very unsatisfying as an answer, I hope that seeing that there are lots of similar examples will make the following analogy make sense.
The inclusions produce a map from the coproduct to the product, in my analogy, I want to think that this is like proving an inequality. And sometimes this inequality is a strict inequality (the map is not an isomorphism), but sometimes in special cases, the inequality is an equality (the map is an isomorphism), and then something special and interesting happens. But the point is that asking why something is a strict inequality is hard to answer with anything other than because we can prove that they aren't the same. Instead, asking when do we have equality may be more fruitful. (Not a perfect analogy, granted.)
Best Answer
A coproduct of the family $\{G_i\}_{i\in I}$ is a group $C$ together with maps (embeddings in fact, though that can be deduced from the universal property) $\lambda_i\colon G_i\to C$ with the following universal property:
Lang is noting that you can embedd each $G_i$ into the product using the functions $\lambda_i$. He also notes that if $i\neq j$, then for every $x\in G_i$, $y\in G_j$, you have that $\lambda_i(x)$ and $\lambda_j(y)$ commute. That means that the image of $\lambda_i(x)$ and the image of $\lambda_j(y)$ will commute under any homomorphism $F\colon\prod G_i\to H$.
But this is a problem: if you can find a group $H$ and morphisms $f_i\colon G_i\to H$ with the property that there are $i$ and $j$, $i\neq j$, such that not every element of $f_i(G_i)$ commutes with every element of $f_j(G_j)$, then there is no way to find a morphism $F\colon \prod G_i\to H$ that will satisfy $F\circ\lambda_i=f_i$. And such groups and maps abound, which means that the fact that images of the $G_i$ inside the product commute with one another doom the product from being able to be the coproduct in the category of all groups.
(The product does have the relevant universal property relative to any family of morphisms $f_i\colon G_i\to H$ that have the additional property that every element of $f_i(G_i)$ commutes with every element of $f_j(G_j)$ whenever $i\neq j$; and of course you can see why, at least in the finite index case, the product will work as a coproduct for abelian groups, that is, if every group in sight is abelian.)