= Solution
Give $\mathcal O(U)$ the <compact-open topology>. Choose a finite polygonal cycle $\Gamma$ in $U\setminus\sigma_A(x)$ whose <winding number> is one on the <Banach algebra spectrum> and zero outside $U$. One can take the oriented <boundary> of a finite union of small squares surrounding the <compact> <Banach algebra spectrum> and contained in $U$. Define
$$
\Theta_x(f)=\frac1{2\pi i}\int_\Gamma f(z)(z1-x)^{-1}\,dz.
$$
The integral is an algebra-norm integral of a <continuous> function on finitely many segments. Cauchy's theorem, applied after any <bounded linear functional> on $A$, makes it independent of the cycle. Moreover
$$
\|\Theta_x(f)\|\le C_\Gamma\sup_{z\in\Gamma}|f(z)|,\qquad
C_\Gamma=\frac{\operatorname{length}(\Gamma)}{2\pi}\max_\Gamma\|(z1-x)^{-1}\|,
$$
so it is <continuous> for the <compact-open topology>.
Here is a direct verification of the algebra properties and uniqueness using the allowed <Runge theorem>. The version required is that every <holomorphic function> on an <open> plane <set> is a locally uniform limit of <rational functions> with poles outside that <set>, allowing a <polynomial> part. A <rational function> $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x)^{-1}$, since the roots of $q$ avoid the <Banach algebra spectrum>. These rational values form a <unital algebra> <homomorphism>.
They agree with the <contour integral>. First $\frac1{2\pi i}\int_\Gamma(z1-x)^{-1}\,dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^j1-x^j=(z1-x)\sum_{k=0}^{j-1}z^{j-1-k}x^k$ then proves the assertion for <polynomials>. For $a\notin U$, the <resolvent identity> gives
$$
\frac{(z1-x)^{-1}}{z-a}
=(x-a1)^{-1}\left((z1-x)^{-1}-\frac1{z-a}1\right).
$$
The cycle has <winding number> zero at $a$, so integration gives $(x-a1)^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the <Banach algebra spectrum>. Differentiating with respect to $a$ gives the analogous result for every power $(z-a)^{-k}$, and partial fractions give it for all the <rational functions> under consideration.
If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_ns_n\to fg$ locally uniformly and multiplication in $A$ is <continuous>, their limits satisfy $\Theta_x(fg)=\Theta_x(f)\Theta_x(g)$. Also $\Theta_x(1)=1$ and $\Theta_x(Z)=x$. Conversely any <continuous> <unital> complex-algebra <homomorphism> sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on <rational functions> and then, by Runge density and <continuity>, on all of $\mathcal O(U)$. This proves \b[existence and uniqueness of the <holomorphic functional calculus>].
For its spectral mapping property, first note compatibility on a smaller <open> spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu\notin g(\sigma_A(x))$, choose such a $W$ where $\mu-g$ never vanishes. Then
$$
(\mu1-\Theta_x(g))^{-1}=\Theta_x\bigl((\mu-g)^{-1}\bigr),
$$
with the right side evaluated on $W$. Conversely, for $\lambda\in\sigma_A(x)$, write $g(z)-g(\lambda)=(z-\lambda)q(z)$ with $q$ <holomorphic> on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda\in\sigma_A(x)$. Thus
$$
\boxed{\sigma_A(\Theta_x(g))=g(\sigma_A(x))\subseteq g(U)\subseteq V.}
$$
Pullback $h\mapsto h\circ g$ is <continuous> from $\mathcal O(V)$ to $\mathcal O(U)$ because the image under $g$ of a <compact> <set> is <compact>. The composite $h\mapsto\Theta_x(h\circ g)$ is a <continuous> <unital> <homomorphism> taking the coordinate function to $y=\Theta_x(g)$. Uniqueness of the calculus at $y$ proves
$$
\boxed{\Theta_y(h)=\Theta_x(h\circ g).}
$$
For the exponential conclusion, let $K=\sigma_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a <simply connected> <open> neighborhood of $K$ avoiding zero, and hence admits a <holomorphic logarithm> $\ell$. <Set> $y=\Theta_x(\ell)$. The composition rule gives
$$
\boxed{e^y=\Theta_x(e^\ell)=\Theta_x(Z)=x.}
$$
The calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible <matrix> over the <complex numbers> has a finite <Banach algebra spectrum> avoiding zero. Removing finitely many plane points leaves a <connected> <set>, so the hypothesis always holds: \b[every invertible <matrix> over the <complex numbers> is an exponential of a <matrix> over the <complex numbers>].
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