My question relates to reconciling the definition of the spectrum $\sigma(a)$ of a point $a$ in a $C^*$-algebra $A$ in both the unital case and the more general case using the unitisation, since I appear to have a mistake in my understanding. I outline below the definitions I am using and the problem I have arrived at.
Each $C^*$-algebra $A$ has a unitisation $\tilde{A}$, which is defined to be the complex vector space $A\times\mathbb{C}$ with the multiplication $$(a,\lambda)(b,\mu):=(ab+\lambda b+\mu a,\lambda\mu),$$ the involution $(a,\lambda)^*:=(a^*,\overline{\lambda})$ and the norm $$\|(a,\lambda)\|:=\max\lbrace\|L_{a,\lambda}\|,|\lambda|\rbrace,$$ where $L_{a,\lambda}:A\to A$ is given by $L_{a,\lambda}(b):=ab+\lambda b$. $A$ then sits inside $\tilde{A}$ as a maximal ideal.
If $A$ is unital, we define its spectrum at $a$ to be $$\sigma(a):=\lbrace\lambda\in\mathbb{C}:\lambda 1_A-a\notin\text{Inv}(A)\rbrace.$$ Then, in general, we define $$\sigma_A(a):=\sigma_{\tilde{A}}((a,0))\equiv\lbrace\lambda\in\mathbb{C}:\lambda(0_A,1)-(a,0)\notin\text{Inv}(\tilde{A})\rbrace.$$ First of all, $\sigma(1_A)=\lbrace1\rbrace$ based on the definition for unital $C^*$-algebras. Now suppose that $0\notin\sigma_{\tilde{A}}((1_A,0))$. Then $(1_A,0)\in\text{Inv}(\tilde{A})$ so there exists $b\in A$ and $\mu\in\mathbb{C}$ for which $(1_A,0)(b,\mu)=(b+\mu1_A,0)=(0_A,1)$, a contradiction, so $0\in\sigma_A(1_A)$ based on the second definition!? Where have I gone wrong?
The mistake is to assume that the unitization process works fine for a unital algebra. It "works", but not as you seem to think it does.
Of course you can construct the unitization for unital $A$, but what you get is a different object. Similar to what you say, $(1_A,0)(0,1)=(1_A,0)$, showing that $(1_A,0)$ is not the unit of the unitization. And, as you say, $(1_A,0)$ is not invertible.
Whether $A$ is unital or not, the unitization is a new algebra, that contains $A$ as an essential ideal. In particular, when you start with unital $A$ the spectrum of $1_A$ changes as by construction all original elements of $A$ have zero in their spectrum as elements of $\tilde A$.