Suppose that $R$ is a ring, not necessarily commutative nor associative. Assume that for every non-zero $a \in R$, the left multiplication map
$$ \lambda\_a \colon R \to R \colon x \mapsto ax $$
is invertible. (We do *not* assume that its inverse is again a left multiplication map $\lambda\_b$ for some $b \in R$.)


Is such a ring $R$ necessarily unital (i.e., does it have a unit $1 \in R$), and is it then a division ring (i.e., does every element $a$ have a two-sided inverse)?


Note: *if* $R$ is unital and associative and every $\lambda\_a$ ($a \neq 0$) is invertible, then it is indeed a division ring, and the inverse of $\lambda\_a$ is equal to $\lambda\_b$ where $b=a^{-1}$.



Invertibility of all left multiplication maps in non-unital rings
What if your ring is the $\mathbf{R}$-algebra $\mathbf{R}^2$ with the bilinear law $$(x,y)(z,t)=\begin{pmatrix}2x & -y \\ y & x\end{pmatrix}\begin{pmatrix}z \\ t\end{pmatrix}=(2xz-yt,yz+xt)\quad?$$ It's even commutative.


