Let $B_i\in\mathbb{B}_i$ be nonempty for each $i\in I$, and consider the set $I_0 = \{i\in I\mid B_i\neq E_i\}$. If $I_0$ is countable, then $\prod B_i\in\prod\mathbb{B}_i$. If the set $I_0$ is uncountable, then $\prod B_i\notin\prod\mathbb{B}_i$. In particular, if $I$ is countable then you get all the "box" sets, and if it is uncountable, you will never have a set $\prod B_i$ where $\emptyset\neq B_i\neq E_i$ for all $i\in I$.
The key is that the product $\sigma$-algebra is the smallest $\sigma$-algebra that makes all the canonical projections measurable; that is, such that for every $i\in I$, every $B_i\mathbb{B}_i$, $\pi_i^{-1}(B_i)\in\prod\mathbb{B}_i$.
Let $X=\prod\limits_{i\in I}E_i$, and $\mathbb{B}=\prod\limits_{i\in I}\mathbb{B}_i$, for simplicity.
First, assume that $I_0$ is countable, and let $B_i\in\mathbb{B}_i$ for each $i$; I claim that $\prod\limits{i\in I}B_i\in\mathbb{B}$.
To see this, for each $i\in I_0$ let $S_i =\pi_i^{-1}(E_i-B_i)= (E_i-B_i)\times\prod\limits_{j\neq i}E_j$. Then $S_i\in\mathbb{B}$, and hence so is $\mathop{\cup}\limits_{i\in I_0}S_i$. Since $I_0$ is countable, this is an element of $\mathbb{B}$, being a countable union of elements of $\mathbb{B}$. An element $(e_i)\in X$ lies in $\cup S_i$ if and only if $e_i\notin B_i$ for some $i\in I_0$. Therefore, $(e_i)\in X-\cup S_i$ if and only if $e_i\in B_i$ for all $i\in I_0$. That is, $X-\cup S_i = \mathop{\prod}\limits_{i\in I_0}B_i\times\mathop{\prod}\limits_{i\notin I_0}E_i = \prod\limits_{i\in I}B_i$. This is the complement of an element of $\mathbb{B}$, hence lies in $\mathbb{B}$.
Note however that $\mathbb{B}$ will usually contain other stuff besides these sets. Even in the case where $I=\{1,2\}$, you usually have more than just the "box" sets in the $\sigma$ algebra. For example, the product of the Borel $\sigma$-algebra on $[0,1]$ with itself contains the set $([\frac{1}{4},\frac{1}{2}]\times[\frac{1}{4},\frac{1}{2}])\cup([\frac{3}{4},1]\times[\frac{3}{4},1])$, which is not of the form $A\times B$ with $A$ and $B$ Borel subsets of $[0,1]$.
Now assume that $I_0$ is uncountable. I will construct a $\sigma$-algebra that necessarily contains $\mathbb{B}$ (but need not be equal to it), and which does not contain any set of the form $\prod\limits{i\in I}B_i$ in which the subset $I_0 = \{i\in I\mid B_i\neq E_i\}$ is uncountable (in particular, when it is equal to all of $I$).
Let $S$ be the $\sigma$-algebra on $\prod\limits_{i\in I}E_i$ that consists of all sets $A$ of the following form: there exists a countable set $J\subseteq I$, and an element $B\in \mathcal{P}\left(\prod\limits_{j\in J}E_i\right)$, such that $A = B\times\prod\limits_{i\notin J}E_i.$ That is, $A$ is an arbitrary thing in the $J$-coordinates, but is all of $E_i$ in the non-$J$ coordinates.
I claim that $S$ is a $\sigma$-algebra on $X$. It contains the empty set by selecting $B=\emptyset$. It is closed under complements, as $X-A = \left((\prod\limits_{j\in J}E_j) - B\right)\times\prod\limits_{i\notin J}E_i$ is of the same form. And it is closed under countable unions: if $A_n$ is a subset of the desired form, with associated countable $J_n\subseteq I$, letting $J=\cup J_n$ we have that $\cup A_n$ is of the form $C\times\prod\limits_{i\notin J}E_i$ with $C$ a subset of $\prod_{j\in J}E_j$. Since each $J_n$ is countable, $J$ is countable, so this set is in $S$. Thus, $S$ is a $\sigma$-algebra on $X$.
Note also that $S$ contains the $\sigma$-algebra $\mathbb{B}$, since the generators of $\mathbb{B}$ all lie in $S$.
However, if $B_i\in\mathbb{B}_i$ for each $i$, and the set $J=\{i\in I\mid B_i\neq E_i\}$ is uncountable, then $\prod B_i\notin S$, and hence does not lie in $\mathbb{B}$ either. Thus, $\mathbb{B}$ does not contain any set which is a product of elements of the different $\mathbb{B}_i$ in which uncountably many components are not the entire space.
Best Answer
A $\sigma$-algebra on a set $X$ is a collection $\mathcal{A}$ of subsets of $X$ which is closed under countable unions and complements. So, when the article refers to a $\sigma$-algebra on $G$, it means a $\sigma$-algebra on the set $X=G$. The elements of this $\sigma$-algebra are subsets of $G$ with their usual Boolean operations, not elements of $G$.
I would add that this statement in no way at all asserts that such a $\sigma$-algebra exists for arbitrary $G$ (though it turns out that it does). It is just saying that given a $\sigma$-algebra $\mathcal{G}$ on $G$, then the triple $(G,\circ,\mathcal{G})$ is called a measurable group if certain conditions hold.