If $G$ is a group scheme, it's natural to study it's closed points $Cl(G)$. If $G$ is finite type over a field, then from the general fact that morphisms between algebraic varieties preserve closed points we know $Cl(G)$ is closed under group operation hence is a group.
What about the general case? Will $Cl(G)$ be not closed under group operation?
Even for varieties over a field, $Cl(G)$ may not be a group. The problem is that $Cl(G\times G)$ is not the same as $Cl(G)\times Cl(G)$. The group operation gives a map $Cl(G\times G)\to Cl(G)$, but not a map $Cl(G)\times Cl(G)\to Cl(G)$.
For instance, consider the additive group $\mathbb{G}_a=\mathbb{A}^1$ over $\mathbb{R}$. The closed points $Cl(\mathbb{G}_a)$ are equivalence classes of complex numbers under conjugation, which do not have any natural group structure under addition. In particular, if you take two points $p$ and $q$ with residue field $\mathbb{C}$, there is no canonical closed point of $Cl(\mathbb{G}_a\times\mathbb{G}_a)$ which corresponds to the ordered pair $(p,q)$, and so you cannot define a sum of $p$ and $q$.
(In fact, there are two different closed points of $\mathbb{G}_a\times\mathbb{G}_a$ whose projections are $p$ and $q$, coming from the two different maximal ideals of $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}$. The images of these closed points under the addition map $\mathbb{G}_a\times\mathbb{G}_a\to\mathbb{G}_a$ correspond to the two different ways to "add" the equivalence classes of complex numbers that $p$ and $q$ represent. That is, if $p$ corresponds to the complex numbers $a\pm ib$ and $q$ corresponds to the complex numbers $c\pm id$, the two different possible sums are $(a+b)\pm i(c+d)$ and $(a+b)\pm i(c-d)$.)