Let $k$ be a finite field and $A$ be a finite $k$-algebra. Assume we are given zero divisor $x$ in $A$ and consider the left ideal $Ax$. In (a), Rónyai assumes that one can find a right identity $r$ of $Ax$. Of course, this would exist if we could solve $a_ie=a_i$ for a basis $\{a_i\}$ of $A$, and if we express $e$ as $e=\sum_j\varepsilon_j a_j$, this amounts to solving the system of $n^2$ linear equations $\sum_j a_i a_j \lambda_j = a_i$.
However, I do not see why this system should have a solution. The "cube" $(a_i a_j)_{ij}$ is certainly singular since $Ax$ contains zero divisors.
(a) Ronyai: Computing the structure of finite algebras, after lemma 5.1.
The context is a lot more specific than you initially mentioned. In the passage you are citing, we are talking about matrix algebras over fields.
It is well-known that every left ideal $L$ of a matrix algebra is generated by an idempotent $e$ (that is, $L=Re$), and that idempotent acts as a right identity of the left ideal, since $(Re)e=Ree=Re$.
That all left ideals are generated by idempotents is an easy consequence of the fact that the matrix ring has identity and that every left ideal is a direct summand.
The statement does not hold for finite algebras in general. For example, take $F_2[x]/(x^3)$ and let $A=(x)/(x^3)$ and the ideal $(x^2)/(x^3)$ has no left or right identity.