Suppose $E_1, E_2, E_3$ are 4 by 4 identity matrices, except $E_1$ has $a, b, c$ in column 1
and $E_2$ has $d, e$ in column $2$ and $E_3$ has $f$ in column 3 (below the $1$ s). Multiply $L = E_1E_2E_3$ to show that all these nonzeros are copied into $L$. $L$ is in the opposite order from elimination (because E3 is acting first) but it's in the correct order to invert elimination and recover A.
$\Large{{1.}}$ I verified the product above by multiplying all $3$ matrices. Nonetheless, I'd like to intuit/perceive why $A(i,j)B(i,j) = AB(i,j)$ even after matrix multiplication? I'd be grateful for pictures.
$\Large{{2.}}$ In the main, when does $A_1(i,j)...A_n(i,j) = \Pi_{1 \le i \le n} \; A_i(i,j)$ ensuing matrix multiplication?
For the first question, probably the simplest way to understand this in a general setting is to use induction.
Set your $E$ to be $$ E=\left(\begin{array}{c|c}1 & 0 \\ \hline e & I\end{array}\right)$$ where $e$ is a column vector and $I$ is an identity matrix of the same dimension. Now if $$ L=\left(\begin{array}{c|c}1 & 0 \\ \hline 0 & M\end{array}\right) $$ is a matrix partitioned consistently as $E$, then $$ EL=\left(\begin{array}{c|c}1 & 0 \\ \hline e & M\end{array}\right) $$ That is, the multiplication with $E$ does not change the trailing block $M$ of the matrix $L$ and simply adds the column $e$ of $E$ to the corresponding block of $L$.