Some of what I have read recently on using functional calculus for non-unital (or just arbitrary) $C^*$-algebras has left me puzzled and made me wonder if I have been harbouring a few misconceptions. I outline here my current (possibly incorrect) approach and am hoping to become more confident applying this and clear up any issues. My main source is $\S$ 2.2 of these online notes; a more detailed account of the functional calculus is given here. Apologies for the length.
If $A$ is a unital $C^*$-algebra and $a\in A$ is normal, then there exists a unique isometric $*$-isomorphism $\Psi:C(\sigma(a))\to C^*(\lbrace 1_A,a\rbrace)$ with $\Psi(\text{id})=a$, where $\text{id}:\sigma(a)\to\mathbb{C}$ is given by $\text{id}(\lambda):=\lambda$. This isomorphism is constructed using the Abstract Spectral Theorem and the $*$-isomorphism induced by the evaluation $h\mapsto h(a):\Delta\to\sigma(a)$. Uniqueness follows easily from the fact that the polynomials are dense in $C(\sigma(a))$.
Moving on to the general case, since $\tilde{A}$ is unital and $a$ is normal in $\tilde{A}$, we have a unique $*$-isomorphism $$\Psi:C(\sigma_A(a)\cup\lbrace0\rbrace)\to C^*(\lbrace1_{\tilde{A}},a\rbrace)$$ with $\Psi(\text{id})=a$, where $\text{id}$ is defined on $\sigma_A(a)\cup\lbrace0\rbrace$ this time. Defining $\mathcal{C}(a)$ to be the $C^*$-subalgebra of $C(\sigma_A(a))$ consisting of those $f$ for which there exists a (necessarily unique) continuous extension $\tilde{f}$ to $\sigma_A(a)\cup\lbrace0\rbrace$ with $\tilde{f}(0)=0$, $\Psi$ restricts to a $*$-isomorphism $$\mathcal{C}(a)\to C^*(\lbrace a\rbrace),$$ still mapping $\text{id}$ to $a$ of course but with the constant polynomials cut out (identify $\mathcal{C}(a)$ with those $f\in C(\sigma_A(a)\cup\lbrace0\rbrace)$ with $f(0)=0$) . If $0\notin\sigma_A(a)$, then it follows from the Pasting Lemma that $\mathcal{C}(a)=C(\sigma_A(a))$.
Would it then suffice to use this trimmed down version of the functional calculus unless constant terms are needed in the unital case e.g. $f(a)=1_A-a$?