Let $\rho$ be a state on a nonunital $C^\ast$-algebra $\cal{A}$, $a\in\cal{A}$ and $\rho(a^\ast a)\neq0$. Define $\rho_a:b\mapsto\rho(a^\ast ba)/\rho(a^\ast a)$. Then $\rho_a$ is a state on $\cal{A}$.
It is clear that $\rho_a$ is a positive linear functional and not difficult to show $\rho_a$ is a state if $\cal{A}$ has an identity. But if $\cal{A}$ has no identity, I don't know how to prove $\|\rho_a\|=1$.
Thanks a lot!
Your C$^*$-algebra has an approximate unit $\{e_j\}$, with $\|e_j\|\leq1$ for all $j$. Then $a^*e_ja\to a^*a$. As $\rho_a$ is positive, it is bounded (or you can show that it is bounded by using that $\rho$ is bounded).
Then $$ \rho_a(e_j)=\frac{\rho(a^*e_ja)}{\rho(a^*a)}\to\frac{\rho(a^*a)}{\rho(a^*a)}=1. $$
For the reverse inequality, we use Kadison's Schwarz inequality and the inequality $a^*b^*ba\leq\|b\|^2\,a^*a$ (from $b^*b\leq\|b\|^2$): $$ |\rho_a(b)|^2\leq\rho_a(b^*b)=\frac{\rho(a^*b^*ba)}{\rho(a^*a)}\leq\|b\|^2\,\frac{\rho(a^*a)}{\rho(a^*a)}=\|b\|^2. $$