= Solution
Work over the complex scalar field. If $\lambda\in\sigma_A(x)$, the <ideal> generated by $x-\lambda1$ is proper: otherwise commutativity would make $x-\lambda1$ invertible. By <Zorn lemma> extend it to a <maximal ideal> $M$. This <ideal> is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper <ideal> cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline M=M$.
The quotient $A/M$ is a complex unital <Banach algebra> and a division algebra. The <Gelfand-Mazur theorem> identifies it with $\mathbb C$, giving a <character of an algebra> $\varphi$ with $\varphi(x)=\lambda$. Conversely, the argument in (a) shows that every $\varphi(x)$ is in the spectrum. Thus
$$
\boxed{\sigma_A(x)=\{\varphi(x):\varphi\in\Phi_A\}.}
$$
Here, as usual, a unital <Banach algebra> has a nonzero identity; the zero algebra is excluded.
The <holomorphic functional calculus> assigns to every function holomorphic on an open neighbourhood $U$ of $\sigma_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of <rational functions> having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula is
$$
\boxed{f(x)=\frac1{2\pi i}\int_\Gamma f(z)(z1-x)^{-1}\,dz.}
$$
Here $\Gamma$ is a finite oriented contour cycle in $U\setminus\sigma_A(x)$, with <winding number> one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the <resolvent of an element>.
For the exponential, use a circle $|z|=R>\|x\|$. The uniformly convergent Neumann expansion $(z1-x)^{-1}=\sum_{n\ge0}x^n z^{-n-1}$ may be integrated term by term. The scalar <Cauchy integral formula> gives $\frac1{2\pi i}\int e^z z^{-n-1}\,dz=1/n!$, proving
$$
\boxed{\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}.}
$$
This series converges absolutely in the <Banach algebra> <norm>.
A <character of an algebra> is bounded and takes the resolvent to $(z-\varphi(x))^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\varphi(f(x))=f(\varphi(x))$. The <character of an algebra> description of the spectrum, now applied to $f(x)$, proves the <spectral mapping theorem>:
$$
\boxed{\sigma_A(f(x))=f(\sigma_A(x)).}
$$
In particular this spectrum is contained in $V$ when $V\supset f(U)$.
For the <composition rule for holomorphic functional calculus>, choose a relatively compact open neighbourhood $D$ of $f(\sigma_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with <winding number> one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of $\sigma_A(x)$ with $\overline\Omega\subset U$ and $f(\overline\Omega)\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta\in\Delta$ the function $h_\zeta(z)=(\zeta-f(z))^{-1}$ is holomorphic near $\overline\Omega$. Locality and multiplicativity of the calculus give
$$
(\zeta1-f(x))^{-1}=h_\zeta(x)
=\frac1{2\pi i}\int_\Gamma\frac{(z1-x)^{-1}}{\zeta-f(z)}\,dz.
$$
Insert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, so
$$
\begin{aligned}
g(f(x))&=\frac1{(2\pi i)^2}\int_\Gamma (z1-x)^{-1}
\left(\int_\Delta\frac{g(\zeta)}{\zeta-f(z)}\,d\zeta\right)dz\\
&=\frac1{2\pi i}\int_\Gamma g(f(z))(z1-x)^{-1}\,dz
=\boxed{(g\circ f)(x)}.
\end{aligned}
$$
This proves equality as algebra elements. Equality of their <character of an algebra> values alone would be insufficient in a <Banach algebra> with nonzero radical.
Finally, $\|x\|<1$ places the spectrum in the open unit disk. On that disk $f(z)=\log(1-z)=-\sum_{n\ge1}z^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-\sum_{n\ge1}x^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\|x\|<R<1$. Since $\exp(f(z))=1-z$, the composition rule gives the <logarithm of an element near the identity> formula
$$
\boxed{1-x=\exp(y),\qquad y=-\sum_{n=1}^\infty\frac{x^n}{n}.}
$$
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