Ignoring the P.S. the answer is an easy no: $A$ could be any unital $*$-algebra with a codimension $1$ $*$-ideal $C$. Then let $B = \mathbb{C}\cdot e$ where $e$ is the unit of $A$, so a $B$-$B$ bimodule is just a vector space. Of course $B$ is endowed with a C*-norm but $A$ is arbitrary so it need not have a C*-norm. The P.S. doesn't add anything because the C*-norm on $B$, if it exists, is unique, so if $A$ has a C*-norm then its restriction to $B$ must be $\|\cdot\|$.
Edit: if the C*-norm on $B$ is only a pre-C*-norm, as per the edit, then it need not be unique, but the answer is still no. Let $A_0$ be the polynomials in $x$ with norm inherited from $C[0,1]$ and let $A = A_0 \oplus \mathbb{C}$ ($l^\infty$ direct sum). Let $f = (f_0, 2)$ where $f_0$ is the polynomial $x \mapsto x$, let $B$ be the set of unital polynomials in $f$, and let $C$ be the $\mathbb{C}$ summand. Then $B$ is isomorphic to $A_0$ as $*$-algebras, but the norm on $A_0$ transferred to $B$ has no extension to $A$. That's because $f -2$ is not invertible in $A$ so the norm of $f$ in $A$ must be at least $2$, whereas its norm in $B$ is $1$.
