Why, if principal ideals are different, they are not 2-sided? There is an argument about ring ideals which I do not understand.

Let $R$ be a ring with $1$,
  $$a=\begin{bmatrix}1&0\\0&0\end{bmatrix},$$ $I=\{a\}\cdot M_2(R)$,
  $J=M_2(R)\cdot\{a\}$. $I$ is a right ideal and $J$ is a left ideal of
  $M_2(R)$. Observe that, if $R$ is not trivial, $I\not=J$ and hence
  none of them is 2-sided.

I do not understand this “hence”. How is $I\not=J$ relevant? I can prove that $J$ is not a right ideal in a concrete way: the right column of every member of $J$ is zero and $$\begin{bmatrix}1&0\\0&0\end{bmatrix}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}
= \begin{bmatrix}0&1\\0&0\end{bmatrix}.$$
I guess I need to prove something like “if $J=S\cdot\{a\} $ is a right ideal, then $J=\{a\}\cdot S $” for an abstract ring $S$ with $1$. I proved the following. $1\in S$, $1\cdot a\in J $, $a\in J $. [Let $s\in S$ be given. $a\cdot s\in J $ because $J$ is a right ideal.] Hence $\{a\}\cdot S\subseteq J $. What next?
[Update 2017-09-12 19:48:28+03:00. It seems that this is a mistake in the textbook. I publish the parameters of the textbook so the authors can fix it.
Menini, Claudia, and Freddy Van Oystaeyen. Abstract Algebra: A Comprehensive Treatment. Marcel Dekker, 2004.]
 A: You're right to be uneasy about the wording: it is not completely correct.
It seems to be saying "for a ring $S$, if $xS\neq Sx$, then $xS$ and $Sx$ are both not two-sided ideals." However, there exist examples of rings such that $xS$ is a two-sided ideal, and yet $Sx$ is not, and in such a case, $xS\neq Sx$.
Here is a concrete example (entry at DaRT). Take any field endomorphism $\sigma:F\to F$ such that $[F:\sigma(F)]>1$. Take the twisted polynomial ring $F[x;\sigma]$ where $x\lambda:=\sigma(\lambda)x$ for each $\lambda \in F$, and let $R$ be the quotient ring $F[x;\sigma]/(x^2)$.  Then $xF[x;\sigma]=xF=\sigma(F)x$ is a principal right ideal which is not a principal left ideal, and $F[x;\sigma]x=Fx$ is a two-sided principal ideal.
What we can say is that if $x\in S$, then $xS$ would be an ideal iff $Sx\subseteq xS$. Similarly, $Sx$ would be an ideal iff $xS\subseteq Sx$.
In the example given, it would be better to say that you can show that neither $aS$ nor $Sa$ are two-sided ideals, because neither one contains the other one.
