Product of principal ideals: $(a)\cdot (b) = (a b)$ In which kinds of rings $R$ does the following hold:
$$(a)\cdot (b) = (ab) \; ?$$
With $a, b\in R$, $(a)$ denoting the (two-sided) ideal generated by $a$ and the multiplication of ideals $I, J\subset R$ defined as 
$$ I\cdot J = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I, y_i \in J \biggr\}\, .$$
It seems to me that it only holds for commutative rings with $1$. Is that right?

Ok, I'm trying a proof:


Let $R$ be commutative with $1\in R$. Then $(a)\cdot (b) = (a b)$ for any $a, b\in R$.

First let $I_a := \{ra : r\in R\}$, we're going to show that 
$$(a) = I_a\, .$$
$I_a$ is obviously an ideal. Since $1 \in R$ we have $1 \cdot a \in I_a$, so $(a) \subset I_a$. On the other hand, any $x \in I_a$ can be written as $x = ra$ and so must be an element of $(a)$. This proves that $(a) = I_a$.
Then $$(a)\cdot (b) = I_a \cdot I_b = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I_a, y_i \in I_b \biggr\}  = \biggl\{\sum_{i=1}^n (r_i a)  (s_i b) : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b \sum_{i=1}^n r_i s_i : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b r : r \in R \biggr\} = I_{ab} = (ab)\, .$$
We obviously had to use that $R$ is commutative. In the last step we also used that every $r \in R$ can be written as $r=\sum_{i=1}^n r_i s_i$. This is because $1 \in R$, so with $n=1$ we have $r = 1\cdot r$.
 A: Take $R:=M_2(\mathbb{R})$. 
$$a=b:=\begin{pmatrix}0&1\\0&0\end{pmatrix} $$
Then :
$$(ab)=(0)=\{0\} $$
However :
$$\begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}0&1\\1&1\end{pmatrix}=\begin{pmatrix}1&1\\0&0\end{pmatrix} \in (a),(b) $$
And :
$$\begin{pmatrix}1&1\\0&0\end{pmatrix}\begin{pmatrix}1&1\\0&0\end{pmatrix}=\begin{pmatrix}1&1\\0&0\end{pmatrix}\neq 0$$
So $(a).(b)\neq(ab)$.
Here I used the fact that there are nilpotents of order $2$ in the  non-commutative ring. I don't know if you can find an example for a non-commutative-ring with no nilpotent elements (seems harder to me...).
Now for $(a)(b)=(ab)$ in commutative rings.
I claim that this does not hold for commutative ring without unity. First a definition :

In a commutative ring $A$, for $a\in A$, I define $(a)$ to be the set $\{ba|b\in A\}$. It is an ideal of $A$. Note that this definition does not imply that $a\in (a)$ (it does if $A$ has a unity).

Take $A_0:=\mathbb{Z}[X,Y]$ and $A:=(X,Y)$ it is an ideal of $A$ and hence a ring (although it has not a unit, it is an additive group which is closed for the multiplication). From now on any ideal has to be taken as an ideal of the ring $A$ (and not $A_0$).
Then take $a:=X$ and $b:=Y$. I claim that $abX=X^2Y\in (ab)$ but :
$$X^2Y\notin (a)(b) $$
The proof is straightforward because any element of $(a)$ or $(b)$ is either null or has total degree $\geq 2$ hence a non null element of $(a)(b)$ has total degree $\geq 4$ but $X^2Y$ is of total degree $3$. 
So I would conclude by saying that both commutativity and unity are essential to this result. However I am still trying to find a non-commutative ring in which $ab=0$ implies $a=0$ or $b=0$ with a unit for which the relation does not hold... Let's see if this works :
$$A:=\mathbb{C}<X,Y> $$
It is the ring of "polynomials" over $\mathbb{C}$ for which $XY\neq YX$ (it exists...). I claim that this is a ring with a unity (it contains $\mathbb{C}$) and furthermore if $ab=0$ then $a=0$ or $b=0$ for this ring. However if :
$$a:=X^2\text{and } b:=Y^2 $$
Then $X^2Y\in (a)$, $XY^2\in (b)$ but :
$$X^2YXY^2\notin (X^2Y^2) $$
So the relation does not hold for "non-commutative domain" even with unity.
