If $A$ is a commutative Banach algebra and $f$ is holomorphic on $\Omega\subseteq\mathbb{C}$, where $\Omega$ contains $\mathscr{R}(\hat{x})$, I want to prove the existence of a $y\in A$ with $\hat{y}=f\circ\hat{x}$, where $x\in A$.
We have that $\mathscr{R}(\hat{x})=\sigma(x)$ for all $x\in A$. Thus, since $f$ is holomorphic in the open set which contains $\sigma(x)$, it follows that $\Omega\equiv A_{\Omega}$ and $f\equiv\tilde{f}\in \tilde{H}(A_{\Omega})$. Hence $$ \tilde{f}\circ\hat{x}=\tilde{f}(h(x))=\tilde{f}(\sigma(x)), $$ where $h\in\Delta$ is a complex homomorphism.
Since $f:A_{\Omega}\to\mathbb{C}$ and $h:A\to\mathbb{C}$, it follows that $f\circ\hat{x}$ is a complex homomorphism of $A_{\Omega}$. So we can compute that $\|f\circ\hat{x}\|_{\infty}=\max_{h\in\Delta}|f(h(x))|\le\|x\|$ for all $x\in A_{\Omega}$. Then by the Hahn-Banach theorem it follows that $f\circ\hat{x}$ extends to a complex linear functional $\phi$ on $A$ that satisfies $|\phi(x)|\le\|x\|$ for all $x\in A$.
I still have to show that the range of $\phi(x)$ is $\sigma(f(x))$, and that $\phi$ is a homomorphism, then $\phi\equiv y$, essentially. But I don't know how to proceed. There is a theorem in Rudin's Functional Analysis (10.29), which I think could be of use and it states that $\tilde{h}=\tilde{g}\circ\tilde{f}$ if $h=g\circ f$. Essentially that was what I was working towards.
I am not aware of how to complete exercise via the path you're following, however there is a fairly simple solution. This will follow the notation of Rudin's Functional Analysis. Using the functional calculus, put $$y=\frac{1}{2\pi i}\int_\Gamma f(\lambda)(\lambda e-x)^{-1}\ d\lambda, $$ where $\Gamma$ surrounds $\sigma(x)$ in $\Omega$. For $h\in\Delta$, we have $$h(y)=\frac{1}{2\pi i}\int_\Gamma f(\lambda)h((\lambda e-x)^{-1})\ d\lambda =\frac{1}{2\pi i}\int_\Gamma\frac{f(\lambda)}{\lambda-h(x) }\ d\lambda =f(h(x)),$$ and therefore $\hat{y}=f\circ\hat{x}$.