Let $S_1\subset B(H_1)$ and $S_2\subset B(H_2)$ be two normed linear spaces, where $H_1$ and $H_2$ are two Hilbert spaces. If $f:S_1\rightarrow S_2$ is a continuous and injective linear map, meanwhile $f$ can be extended to a homomorphism $\overline{f}:C^*(S_1)\rightarrow C^*(S_2)$, where $C^*(S_i)$ are the C*-algebra generated by $S_i$ ($i=1,2$).
Can we get that $\overline{f}$ is injective?
Is this extension $\overline{f}$ always exist?
The answer to both is no.
Take $S_1 = \{\lambda S\ \mid \lambda \in \mathbb{C}\}$ where $S \in B(\ell^2(\mathbb{N}))$ is the unilateral shift and $S_2 = \{\lambda U \mid \lambda \in \mathbb{C}\}$ where $U$ is some unitary with full spectrum. $f$ can be taken to be $f(\lambda S) = \lambda U$ which is clearly continuous, linear, and injective. Now if we look at the C*-algebras generated by $S$ and $U$, $C^*(S_1)$ gives the Toeplitz algebra and $C^*(S_2)$ gives $C(\mathbb{T})$. The former can never inject into the latter as we can't have a non-abelian C*-algebra sitting inside of an abelian one. So this negatively answers your first question, without even considering the existence of such an extension (although the extension actually exists in this case).
For (2) the answer is again no, but lets look to the finite-dimensional case. Take $S_1 = S_2 = M_2$, the 2x2 matrices acting on $\mathbb{C}^2$. Take $f: S_1 \to S_2$ to be the transpose. Its clearly a linear isometry, but not a *-homomorphism.