Is the logarithm always inverse to the exponential in Banach algebras? Let $A$ be a real unital Banach algebra. For $x\in A, r> 0$, denote by $B(x,r)\subset A$ the open ball of radius $r$ centred at $x$. Define $$\log(1+\cdot):B(0,1)\to A,\quad\log(1+x):=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^k.$$
Is it always true that $\log(1+\cdot)$ is a homeomorphism onto its image with inverse $\exp(x):=\sum_0^\infty\frac{x^k}{k!}$? If not, can we perhaps say that $\exists\varepsilon>0$ so that $\log(1+\cdot)\vert_{B(0,\varepsilon)}$ is a homeomorphism onto its image?

In the case where $A=\mathbb R$, this is of course true and is proven here, for example. However, I can't seem to generalise that proof as it uses techniques such as differentiating power series term-by-term and the fact that $(x^k)'=kx^{k-1}$, which doesn't hold in noncommutative Banach algebras.
 A: Let $A,\|.\|$ be a complete normed unital algebra where $\|\frac{x}{n}\|=\frac{\|x\|}{n}$.
$$\exp(x)-1 \overset{def}=  \lim_{N \to \infty}\sum_{n=1}^N \frac{x^n}{n!}, \qquad 
\log(1+x) \overset{def}=  \lim_{K \to \infty}\sum_{k=1}^K \frac{(-1)^{k+1} x^k}{k} $$
For any $x \in A,  \|x\| < 1$ and any $y \in A$
$$\|\exp(y)-1\| \le \sum_{n=1}^\infty \frac{\|y\|^n}{n!} = \exp(\|y\|)-1 < \infty, \\ \|\log(1+x)\| \le \sum_{k=1}^\infty \frac{\|x\|^k}{k} = -\log(1-\|x\|) < \infty \\ \|\exp(\log(1+x))-1\| \le \sum_{n=1}^\infty \frac{\|\sum_{k=1}^\infty \frac{\|x\|^k}{k}\|^n}{n!}=\exp(-\log(1-\|x\|))-1 < \infty$$
Thus everything converges absolutely and everything is continuous. Thus we can change the order of summation to write $$(\log(1+x))^n = \sum_{k=1}^\infty a_{n,k} x^k, \qquad \exp(\log(1+x)) -1= \sum_{n=1}^\infty \frac{1}{n!}\sum_{k=1}^\infty a_{n,k} x^k$$
Changing the order of summation again
$$\exp(\log(1+x))-1 = \sum_{k=1}^\infty x^k \sum_n \frac{a_{n,k}}{n!}= x$$
where $\sum_n \frac{a_{n,k}}{n!} = 0$ for $k > 1$ is a consequence of our knowledge in real/complex analysis.
