The answer is no: unless $G$ is compact, its subspace topology inside $bG$ (called its [Bohr topology](https://www.ams.org/mathscinet-getitem?mr=181700)) is *much weaker* than its own. So $\beta:G\to bG$, while continuous with dense image, is not an [embedding](https://en.wikipedia.org/wiki/Embedding#General_topology) (homeomorphism onto $\beta(G)$), hence not a [compactification](https://en.wikipedia.org/wiki/Compactification_(mathematics)#Definition) in the sense of topologists.

Rudin ([1962](https://www.ams.org/mathscinet-getitem?mr=152834), pp. [30-31](https://books.google.com/books?id=k3RNtFSBH8gC&pg=PA31)) notes this without proof. Later Katznelson (1973) gave a method to show that very “thin” subsets of $G$ are dense in the Bohr topology: e.g. $\mathbf N$ inside $\mathbf Z$, a parabola inside $\smash{\mathbf R^2}$, etc. For more details and references see (with apologies for the plug) [this paper](https://arxiv.org/abs/1211.3783), esp. Lemma 1(4). 

**Note:** I have restricted my answer to abelian $G$; $\beta$ can be defined for any topological group, but your question only makes sense when $\beta$ is injective, i.e. for “maximally almost-periodic” $G$, and these are very nearly the compact $\times$ abelian ones: see Dixmier ([1977](https://www.ams.org/mathscinet-getitem?mr=458185), §§16.1, 16.4).