Say $R$ is a ring with division. Define $\mathcal{M}_2(R)$ to be the set of 2x2 matrices over $R$ with $\mathcal{M}_2(R)$ a ring.
For what $R$ will $\mathcal{M}_2(R)$ ever be a division ring?
It is easy to show that $\mathcal{M}_2(R)$ is an identity ring, if $R$ itself is an identity ring. Take \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} Where $0$ is $R$'s additive identity and $1$ $R$'s multiplicative identity.
However if we take $\mathbb{R}$ as $R$, there will be entries in $\mathcal{M}_2(\mathbb{R})$ that are not invertible, e.g. \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} But $\mathbb{R}$ is a field so it's a division ring.
And so it got me thinking if there are any $R$ out there, that are also division rings, for which $\mathcal{M}_2(R)$ is also a division ring.
Apologies if this is a bit too obvious, or my premise is flawed, I'm only just getting to grips with rings.
The matrix $e=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$ you gave as an example will never be invertible. It satisfies $e^2=e$ and if it were invertible, we could cancel and get $e=Id$, which is false. In fact, $e$ is a zero divisor already since $(I_2-e)e=0$, and neither $e$ nor $I_2-e$ are zero.
Of course, zero divisors are not invertible in nontrivial rings with identity. If $a,b$ are nonzero and $ab=0$ and $a$ is a unit, then $b=0$ (a contradiction) after multiplying with the inverse of $a$ on the left. One could only conclude the assumption that $a$ is a unit is not possible.
So in general, a matrix ring over any ring with identity, with $n > 1$ will never be a division ring.
Another easy zero divisor is this: if you form the matrix that’s $1$ in the upper right hand corner and $0$ elsewhere, it’ll satisfy $A^2=0$.
$M_n(R)$ will be a division ring precisely when $n=1$ and $R$ is a division ring.