Everything is known. In fact as spectra we have canonically $K(\mathsf{finAb}) = \vee_p K(\mathbb{F}_p)$, and the spectra $K(\mathbb{F}_p)$ are identified in the work of Quillen (see e.g. http://www.math.uiuc.edu/K-theory/1006/). In particular on $\pi_0$ we find $K_0(\mathsf{finAb}) = \oplus_p \mathbb{Z}$, agreeing with your claim, and on $\pi_n$ for $n>0$ we find that $K_n(\mathsf{finAb})$ is $0$ for $n$ even and is $\oplus_p \mathbb{Z}/(p^k-1)$ (non-canonically) for $n = 2k-1$.

To justify the claimed equality $K(\mathsf{finAb}) = \vee_p K(\mathbb{F}_p)$, note first that $\mathsf{finAb}$ is the filtered colimit over increasing finite sets of primes $P$ of the variant $\mathsf{finAb}_P$ where only products of $p$-groups for $p \in P$ are allowed; since K-theory commutes with filtered colimits, it then suffices to show that each $K(\mathsf{finAb}_P) = \prod_{p\in P} K(\mathbb{F}_p)$ and that for $P \subseteq P'$ this identification intertwines the inclusion $K(\mathsf{finAb}_P) \to K(\mathsf{finAb}_{P'})$ with the evident map $\prod_{p\in P} K(\mathbb{F}_p) \to \prod_{p\in P'} K(\mathbb{F}_p)$ which is zero outside of $P$.

But $\mathsf{finAb}_P$ is just the product over $p \in P$ of the categories $\mathsf{finAb}_p$, whose K-theory identifies with that of vector spaces over $\mathbb{F}_p$ by Quillen's devissage theorem. And K-theory commutes with finite products, so that's that.

Here I guess I was actually arguing using Quillen's Q-construction instead of Waldhausen's $S_{\bullet}$-construction. Otherwise I'm not sure how to justify the last step, the devissage. Actually I'm sure all of the above is in Quillen's paper on the Q-construction.