The question is:
"Let $\mathcal{T}$ be an additive group topology on a ring $A$, i.e., a topology $\mathcal{T}$ in $A$ such that the operations $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous (considering the product topology in $A\times A$ determined by $\mathcal{T}$). The subset $B$ of $A$, consisting of all $b\in A$ such that $x\mapsto bx$ and $x\mapsto xb$ are continuous at zero, is a subring of $A$; furthermore, if $A$ has an identity, then for every $b\in B$ invertible in $A$ we have $b^{-1}\in B$."
What I have done:
If $b\in B$, then $x\mapsto bx$ is continuous everywhere in $A$. In fact, for $a\in A$, the map $x\mapsto bx$ is the composition of $x\mapsto x-a$ and $x\mapsto bx$ and $x\mapsto x+ba$, because $x=b(x-a)+ba$, then it is easy to see that $x\mapsto bx$ is continuous in $a$. Similarly, the map $x\mapsto xb$ is continuous everywhere in $A$.
Moreover, the maps $x\mapsto 0x$ and $x\mapsto x0$ are constant equal to $0$, so are continuous, so $0\in B$.
If $b,c\in B$, then $x\mapsto (b-c)x$ is the composition of $x\mapsto (bx,cx)$ and $(x-y)\mapsto x-y$, that are continuous maps, so it is continuous, and similarly, the map $x\mapsto x(b-c)$ is continuous, therefore $b-c\in B$.
If $b,c\in B$, then $x\mapsto (bc)x$ is the composition of $x\mapsto cx$ and $x\mapsto bx$, so it is continuous, and similarly $x\mapsto x(bc)$ is continuous, so $bc\in B$.
If $A$ has an identity, then $x\mapsto 1x$ and $x\mapsto x1$ are the identity function, so are continuous, so $1\in B$.
Where I am stuck:
But I cannot have an idea about how to prove that if $b\in B$ is invertible in $A$ then $b^{-1}\in B$.
What I tried:
If $L_b:x\mapsto bx$ and $R_b:x\mapsto xb$ are the translations, and $b$ is invertible, I tried to do the following, because some of these $L_{1+b^{-1}}$ and $L_{1-b^{-1}}$ is continuous, then $L_{B^{-1}}$ is continuous:
$L_{1+b}L_{1+b^{-1}}=L_2+L_b+L_{b^{-1}}=L_{1+b}+L_{1+b^{-1}}$
$L_{1-b}L_{1+b^{-1}}=-L_b+L_{b^{-1}}=-L_{1+b}+L_{1+b^{-1}}=L_{1-b}-L_{1-b^{-1}}$
$L_{1+b}L_{1-b^{-1}}=L_b-L_{b^{-1}}=L_{1+b}-L_{1+b^{-1}}=-L_{1-b}+L_{1-b^{-1}}$
$L_{1-b}L_{1-b^{-1}}=L_2-L_b-L_{b^{-1}}=L_{1-b}+L_{1-b^{-1}}$
I tried to sum some of these identities without much success.
I think I have to interact the left translations with the right translations but I do not know how.
This is good, because there exists a counterexample to this claim. In the counterexample is provided a group topology on $A=\Bbb Q$ such that $2\in B$, but $2^{-1}\not\in B$.