Let $A$ be a $C^*$-algebra, and $\ f:A \to \mathbb C$ a positive linear functional on $A$. Where $\tilde A$ is the unitization of $A$, is there a nice way to extend $f$ to a positive linear functional on $\tilde A$? Is this always possible?
Two approaches I took to this were to set $f(a + \lambda I) = f(a) + \lambda$ (I identified $a \in A$ with its counterpart in $\tilde A$), and $f(a + \lambda I) = f(a)$. I tried to show that $\|f\|_{\tilde A} = 1$ in the first approach, but I got stuck. The second thing that I tried was just computational: $$ f((a + \lambda I)^*(a + \lambda I)) = f(a^*a + \bar\lambda a + \lambda a^*) + |\lambda|^2 $$ This seemed promising, but I'm not sure how to proceed. It may be my naivety thinking, but I seem to doubt there's any hope for the second approach.
Your second approach cannot succeed, because $f(I)=0$; this guarantees that $f$ is not positive unless its zero (you have $|f(a)|\leq f(\|a\|\,I)=0$).
In your first approach, the right extension is $$ f(a+\lambda)=f(a)+\lambda\,\|f\|, $$ and then we may assume without loss of generality that $\|f\|=1$ (that is, $f$ is a state).
Now we can check positivity directly: if $b\in \bar A$ is positive, then $b=(a+\lambda)^*(a+\lambda)$ for some $a\in A$, $\lambda\in\mathbb C$. Then, using Kadison's inequality, \begin{align} f(b)&=f((a+\lambda)^*(a+\lambda))=f(a^*a+2\text{Re}\,\bar\lambda a+|\lambda|^2)\\ \ \\ &=f(a^*a)+2\text{Re}\,\bar\lambda f(a)+|\lambda|^2\\ \ \\ &\geq |f(a)|^2+2\text{Re}\,\bar\lambda f(a)+|\lambda|^2\\ \ \\ &=|f(a)+\lambda|^2\geq0. \end{align}