Calculating all products of $n-1$ factors when given $n$ factors
Here is the fastest algorithm. I bet. The idea of the algorithm can be seen from the one-line explanation between step 4 and step 5 below.



Input: $e_1,\cdots,e_n\in E$, where $n\ge 3$.
Output: $p_1, \cdots, p_n$, where $p_j=e_1\cdots e_{j-1}e_{j+1}\cdots e_n$, the product of all input elements except $e_j$.
Procedure:


Allocate array $p_1, p_2,\cdots, p_n$ of $n$ element in $E$.
Let $s=e_1$, $p_1=1$
For $i$ from 2 to $n-1$ inclusive, let $p_i = s$ and $s= se_i$.
Let $p_n=s.$

Now 
$$
\begin{pmatrix}
p_1\\p_2\\ p_3\\p_4\\\cdots\\ p_{n-1}\\ p_n 
\end{pmatrix}
\text{is}
\begin{pmatrix}
 1\\e_1\\e_1e_2\\ e_1e_2e_3\\\cdots\\ e_1e_2\cdots e_{n-2}\\e_1e_2\cdots e_{n-2}e_{n-1}
\end{pmatrix}
$$.   
Let $b=e_n$.
For $i$ from $n-1$ to 2 inclusive downwards, let $p_i=p_ib$ and $b = e_ib$.
Let $p_1=b$.




The number of multiplications performed by the algorithm above is $3(n-2)$. 

I believe that is the tight lower bound. However, I have not been successful proving it even though I have tried a few times. I am still trying from time to time.