Let $c$ denote the inverse of $a+b$, i.e., $ac+bc=ca+cb=1$. Multiplying this by $a$ and $b$ from the left and from the right and using $a^2=b^2=0$, we obtain $abc=a$, $cba=a$, and $cab=b$.

Let us show that $R=aR\oplus bR$. It follows from $ac+bc=1$ that $R=aR+bR$. If $ax=by$, then $bax=0$, implying $0=cbax=ax$ in view of $cba=a$.

The $R$-modules $aR$ and $bR$ are isomorphic. Indeed, the rule $bcx\mapsto ax$, $x\in R$, defines an $R$-module homomorphism $bR=bcR\to aR$ because $bcx=0$ implies $ax=0$ due to $abc=a$. Its kernel vanishes: $ax=0$ implies $abcx=0$ because $abc=a$ and, hence, $0=cabcx=bcx$.

It remains to apply Propositions 5 and 6 on page 52 of "Structure of Rings" by N.Jacobson (1964) and to conclude that $R$ is a ring of $2\times2$ matrices.

**Edit.** By suggestion of Martin Brandenburg, I reproduce the Propositions quite literally:

**Proposition 5** $\dots$ *Conversely, if $\frak A$ is a ring with an identity $1$ and ${\frak A}={\frak I}_1\oplus\dots\oplus{\frak I}_n$ is a direct decomposition of $\frak A$ into right ideals which are isomorphic $\frak A$-modules, then there exists a set of matrix units $\{e_{ij}\mid i,j=1,\dots,n\}$ such that ${\frak I}_j=e_{jj}{\frak A}$.*

**Proposition 6.** *Let $\{e_{ij}\mid i,j=1,\dots,n\}$ be a set of matrix units in a ring $\frak A$ with identity $1$, and $\frak B$ the subring consisting of the elements which commute with the $e_{ij}$, $i,j=1,\dots,n$. Then every element of $\frak A$ can be written in one and only one way as
$\sum b_{ij}e_{ij}$ where $b_{ij}\in{\frak B}$ for all $i$ and $j$. Hence
${\frak A}\cong{\frak B}_n$. The ring $\frak B$ is isomorphic to
$e_{11}{\frak A}e_{11}$.*

**Second thought Edit.** Those Propositions are an old stuff. It is probably better to prove them here in two lines.

For any $R$-module $M$, we have
$\text{End}_R(\underbrace{M\oplus\dots\oplus M}_n)\cong\text{Matr}_n(\text{End}_RM)$.
On the other hand, $\text{End}_RR\cong R$.

**Question.** The problem looks a bit artificial. Is it a known exercise?