$\exp(\ln(x))=x$ and $\ln(\exp(y))=y$. Let $(A,1_A,|\cdot{}|)$ be a unital Banach algebra, for instance $A=M_n(\Bbb R)$ or $M_n(\Bbb C)$. What is the union of all open unit balls $B_{\|\cdot{}\|}$ where $\|\cdot{}\|$ ranges over all (topologically equivalent) norms that 


*

*are submultiplicative? 

*are Banach algebra norms?


In a unital Banach algebra we require $|1_A|=1$ and submultiplicativity: for all $x,y$, $|xy|\leq|x||y|$. Every norm may be rescaled to be submultiplicative : suppose $\|\cdot{}\|$ is a norm on $A$, since multiplication $A\times A\to A$ is (bilinear) continuous, there exists  $C>0$ such that for all $x,y\in A,\|xy\|\leq C\|x\|\|y\|$, then $\|\dot{}\|'=C\|\cdot\|$ is submultiplicative.
The motivation for this question comes from trying to understand the domain where the equations from the proposition below remain true. 

Proposition. Suppose $|\cdot{}|$ is submultiplicative. Then for all $x\in B_{|\cdot{}|}(1_A,1)$ and $y\in B_{|\cdot{}|}(0_A,\ln 2)$
  $$\exp(\ln(x))=x\quad\text{and}\quad \ln(\exp(y))=y\,.$$

The logarithm is defined on $B_{|\cdot{}|}(1_A,1)$ by the absolutely converging series
$$\ln(x)=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}(x-1_A)^n.$$
This power series converges absolutely on the union $\mathcal{O}$ of all the open balls $B_{\|\cdot{}\|}(1_A,1)$ where $\|\cdot{}\|$ is submultiplicative. If we define $\mathcal{U}$ as the union of all $B_{\|\cdot{}\|}(0_A,\ln2)$, again, $\|\cdot{}\|$ is submultiplicative, then for all $x\in\mathcal{O}$ and all $y\in\mathcal{U}$,
$$\exp(\ln(x))=x\quad\text{and}\quad \ln(\exp(y))=y\,.$$
These domains are manifestly translated and rescaled versions of a special open set
$$\mathcal S=\bigcup_{\|\cdot{}\|\text{ submultiplicative}} B_{\|\cdot{}\|}$$
Let us also define $\mathcal B\subset\mathcal S$
$$\mathcal B=\bigcup_{\|\cdot{}\|\text{ Banach algebra norm}} B_{\|\cdot{}\|}$$

Question. Can $\mathcal B,\mathcal S$ be explicitely described? What about the case where $A$ is the algebra of square real or complex matrices, furthermore, what domain do we get if we only consider the subordinate matrix norms?


EDIT. In order to have a chance at the bounty, it is enough to answer one of the following questions:


*

*What is $\mathcal{S}$ for a general Banach algebra?

*What is $\mathcal{B}$ for a general Banach algebra?

*What is $\mathcal{S}$ for a (real or complex) matrix algebra?

*What is $\mathcal{B}$ for a (real or complex) matrix algebra?

*What is the union of all open unit balls for subordinate matrix norms?

 A: Let's work with $A=M_n(\mathbb{C})$. Consider $x\in A$, then

$\|x\|<1$ for some subordinate matrix norm iff $\;\mathrm{Spec}(x)\subset D^{\circ}$

where $D^{\circ}=\lbrace z\in\mathbb{C}\text{ s.t. }|z|<1\rbrace$ is the open unit disc. 
Indeed, if $\|\cdot\|$ is subordinate to some norm $|\cdot|$ on $\mathbb{C}^n$, then for any nonzero eigenvector $v$ of $x$ with eigenvalue $\lambda$, $|\lambda||v|=|x(v)|\leq\| x\||v|$ and the implication $(\Rightarrow)$ follows.
On the other hand, if $\mathrm{Spec}(x)\subset D^{\circ}$, and $\mathcal{E}=(e_1,\dots,e_n)$ is a basis of $\mathbb C^n$ such that
$$\mathrm{Mat}(x\,;\mathcal{E})=\begin{pmatrix}\lambda_1&*&&*\\&\lambda_2\\&&\ddots&*\\&&&\lambda_n\end{pmatrix}$$
is upper triangular and has upper coefficients $|*|<\epsilon$ where $\max_i|\lambda_i|+\frac{n(n-1)}2\epsilon<1$ (such bases always exsist by an easy argument) then an easy calculation shows that $\|x\|<1$ for $\|\cdot\|$ the subordinate norm to the norm $|\cdot|$ on $\mathbb{C}^n$ defined by
$$\left|\sum_i v^ie_i\right|=\sum_i |v^i|$$
Thus the union of all open unit balls of $M_n(\mathbb{C})$ of subordinate matrix norms is the set of all matrices $x$ with $\mathrm{Spec}(x)\subset D^{\circ}$.
