Everything is known. In fact as spectra we have canonically K(finAb) $K(\mathsf{finAb}) = \vee_p K(F_p)K(\mathbb{F}_p)$, and the spectra K(F_p) $K(\mathbb{F}_p)$ are identified in the work of Quillen (see e.g. http://www.math.uiuc.edu/K-theory/1006/). In particular on \pi_0 $\pi_0$ we find K_0(finAb) $K_0(\mathsf{finAb}) = \oplus_p Z, \mathbb{Z}$, agreeing with your claim, and on \pi_n $\pi_n$ for n>0 $n>0$ we find that K_n(finAb) $K_n(\mathsf{finAb})$ is 0 $0$ for n $n$ even and is $\oplus_p \oplus_p Z/(p^k-1) mathbb{Z}/(p^k-1)$ (non-canonically) for $n = 2k-12k-1$.
To justify the claimed equality K(finAb) $K(\mathsf{finAb}) = \vee_p K(F_p)K(\mathbb{F}_p)$, note first that finAb $\mathsf{finAb}$ is the filtered colimit over increasing finite sets of primes P $P$ of the variant finAb_P $\mathsf{finAb}_P$ where only products of p-groups $p$-groups for $p \in P P$ are allowed; since K-theory commutes with filtered colimits, it then suffices to show that each K(finAb_P) $K(\mathsf{finAb}_P) = \prod_{p\in P}K(F_p) P} K(\mathbb{F}_p)$ and that for P inside $P ' \subseteq P'$ this identification intertwines the inclusion K(finAb_P) --> K(finAb_{P'}) $K(\mathsf{finAb}_P) \to K(\mathsf{finAb}_{P'})$ with the evident map \prod_{p\in $\prod_{p\in P} K(F_pK(\mathbb{F}_p) --> \to \prod_{p\in P'} K(F_p) K(\mathbb{F}_p)$ which is zero outside of P.$P$.
But finAb_P $\mathsf{finAb}_P$ is just the product over $p \in P P$ of the categories finAb_p, $\mathsf{finAb}_p$, whose K-theory identifies with that of vector spaces over F_p $\mathbb{F}_p$ 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_\dot-construction. $S_{\bullet}$-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.
