Let $ \mathbb{H}$ be a division algebra of quaternions.
Is there an element in $M ( \mathbb{H} ) $ that cannot be written as the left multiplication map and right multiplication map?
Let $D$ be an algebra. For $a, b \in D $, maps $L_{a}, R_{b} : D \longrightarrow D$ are the left multiplication map and right multiplication map, by
$L_{a} (x) := ax$
$ R_{b} (x) := xb$.
On
If $A$ is a finite dimensional algebra over a field $k$ which is simple (has no proper non-zero bilateral ideals) and central (its center is the one-dimensional subspace spanned by its unit element), then the map $\beta:A\otimes_k A^{op}\to\operatorname{End}_k(A)$ such that $\beta(a\otimes b)=L_a\circ R_b$ is an isomorphism of algebras.
This implies that, as soon as $A$ has dimension larger than $1$, there exist linear maps $f:A\to A$ which cannot be written as a composition $R_a\circ L_b$ for any choice of $a$ and $b$ in $A$. Moreover, one can see that in fact every linear map $A\to A$ is equal to a sum $R_{a_1}\circ L_{b_1}+\cdots+R_{a_d}\circ L_{b_d}$ of $d$ such compositions, with $d=\dim A$, and that $d$ is the smallest number with this property.
The algebra $H$ of quaternions is simple (because it is a division algebra) and you can easily check that it is central. Also, there is an isomorphism $H\cong H^{op}$ (given by "conjugation"), so the above result tells us, in this particular case, that $H\otimes H$ is isomorphic to the endomorphism algebra $\operatorname{End}_k(H)$ of $H$ as a vector space, and that there are elements in $\operatorname{End}_k(H)$ which are not of the form $R_a\circ L_b$.
I presume $M(\Bbb H)$ is the ring of $\Bbb R$-linear maps from $\Bbb H$ to $\Bbb H$. Then the $L_a$ and $R_b$ certainly generate $M(\Bbb H)$ as an $\Bbb R$-algebra. But each $L_a$ and $R_b$ is zero or bijective, so the map $\phi\in M(\Bbb H)$ with $\phi(1)=1$, $\phi(i)=\phi(j)=\phi(k)=0$ is not an $L_a$, nor an $R_b$, nor even an $L_a\circ R_b$.