No that's not possible (except the trivial case). Any $*$-homomorphism between $C^*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the identity map of $A$ seens as a map from $(A,\Vert \cdot \Vert)$ to the completion of $(A,\vert \cdot \vert)$ and conclude that two norm are equal.

More concretely it follows that (for self adjoint element) the norm in a $C^*$-algebra coincide with the spectral radius, which force relations between the two norm (here it forces $\vert \cdot \vert \leqslant \Vert \cdot \Vert$ hence the equality of the two norms).

No that's not possible (except the trivial case). Any $*$-homomorphism between $C^*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the identity map of $A$ seens as a map from $(A,\Vert \cdot \Vert)$ to the completion of $(A,\vert \cdot \vert)$ and conclude that two norm are equal.

More concretely it follows from the fact that (for self adjoint elements) the norm in a $C^*$-algebra coincide with the spectral radius, which force relations between the two norm. Here it forces $\vert \cdot \vert \leqslant \Vert \cdot \Vert$ hence the equality of the two norms.