Consider the following proof from the book "$C^*$-algebras and finite-dimensional approximations":
Why does this proof work in the non-unital case? (see the last line). Maybe we have $$\|a + \lambda \| = \sup_{\varphi \in \mathcal{S}} \{|\varphi(a, \lambda)|\}$$ where the states $\varphi: A \to \mathbb{C}$ have been extended (uniquely) to states on the unitization?
The states of $A$ extend uniquely (as states) to the unitization. The extension is $\tilde\varphi(a+\lambda)=\varphi(a)+\lambda$. It easy to check that such extension is a state.
And you have an extension $\psi$ of $\varphi$ that is a state, then $\psi(1)=1$ (a positive functional always satisfies $\psi(1)=\|\psi\|$). That gives you $$ \psi(a+\lambda)=\psi(a)+\psi(\lambda)=\varphi(a)+\lambda. $$
I don't see a way to go about the proof in an obvious way as Nate and Taka make it sound, but here is my take. Write $n=\|a\|$. Then, with $\tilde\varphi$ the unique extension of $\varphi$ to the unitization as a state, \begin{align} \|a+n\|&=\sup_{\|b\|\leq1}\|ab+nb\| =\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\varphi(ab+nb)|\\[0.3cm] &=\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\tilde\varphi((a+n)b)|\\[0.3cm] &\leq\sup_{\|b\|\leq1}\sup_{\varphi\in S}\tilde\varphi(b^*b)^{1/2}\tilde\varphi((a+n)^2)^{1/2}\\[0.3cm] &\leq\sup_{\varphi\in S}\tilde\varphi(\|a+n\|\,(a+n))^{1/2}\\[0.3cm] &=\|a+n\|^{1/2}\,\sup_{\varphi\in S}\tilde\varphi(a+n)^{1/2}. \end{align} Cancelling and squaring we obtain $$ \|a+n\|\leq\sup_{\varphi\in S}\tilde\varphi(a+n), $$ which is the non-obvious inequality that is needed.