Since $A$ is finite, we the set $\{ b^n \mid n \geq 0\}$ is finite, so there exist $m > n \geq 0$ such that $b^m = b^n$. Take such pairs $(m,n)$ with $n$ minimal. If $n\geq 2$, then $ab^2 = b$ implies $b^{m-1} = bb^{m-2} = ab^2b^{m-2} = ab^m = ab^n = \cdots = b^{n-1}$, which contradicts the minimality of $n$. Hence $n = 0$ or $n=1$. In the case $n=0$, we have $b^m = 1$ for some $m\geq 1$, so $ab^2 = b$ implies $ab = bb^{m-1} = 1$ and hence $b(ab-1) = b0 = 0$, as desired. In the case $n = 1$, we have $b^m = b$ for some $m\geq 1$, so $ab^2 = b$ implies $ab = ab^m = b^{m-1}$ and hence $b(ab-1) = b^m-b = 0$.

Romania National Olympiad 2018/12-01 wrote: Let $A$ be a finite ring and $a,b \in A,$ such that $(ab-1)b=0.$ Prove that $b(ab-1)=0.$ Since $A$ is a finite ring, there exists $i > j \ge 2$ such that $b^i= b^j$. Claim 01. There exists $k \ge 2$ such that $b^k = b$. Proof. We know that $(ab - 1)b = 0$, which means that $ab^2 = b$. Therefore, as $i,j \ge 2$, we must have \[ b^{i - 1} = b \cdot b^{i - 2} = ab^2 \cdot b^{i - 2} = ab^i = ab^j = ab^2 \cdot b^{j - 2} = b \cdot b^{j - 2} = b^{j - 1} \]Let $X$ be the set of natural numbers such that if $\ell \in X$, then there exists $\ell' > \ell$ such that $b^{\ell'} = b^{\ell}$. The above claim implies that if $\ell \in X$ and $\ell \ge 2$, then $\ell - 1 \in X$. Therefore, we could conclude that $1 \in X$, which implies the desired claim. To finish the problem, note that \[ bab = bab^k = b(ab^2)b^{k - 2} = b(b)(b^{k - 2}) = b^k = b \implies b(ab - 1) = 0\]

Note that the given condition implies that $ab^2=b$. It is easy to see that for all positive integers $k$, we have $a^kb^{k+1}=b$. Claim. The set $\{b,b^2,b^3,\ldots\}$ forms a cyclic group under multiplication. Proof. Since $A$ is finite, the set $\{b,b^2,b^3,\ldots\}$ is also finite. It follows that there exists positive integers $\beta_1$ and $\beta_2$ with $\beta_1<\beta_2$ such that $b^{\beta_1}=b^{\beta_2}$. Take $\beta_1$ and $\beta_2$ with $n:=\beta_2-\beta_1$ minimal. Then $b=a^{\beta_1-1}b^{\beta_1}=a^{\beta_1-1}b^{\beta_2}=b^{n+1}$. Let $e_b:=b^n$. By the minimality of $n$, the elements $e_b,b,b^2,\ldots,b^{n-1}$ are distinct. Then the set $\{e_b,b,b^2,\ldots,b^{n-1}\}$ forms a cyclic group of order $n$. $\square$ Thus we have $ab=ab^{n+1}=b^n$ so $b(ab-1)=b^{n+1}-b=0$. $\square$