We know that for C*-algebras, $$ K_0(A\oplus B) \stackrel{(*)}{=} K_0(A)\times K_0(B)$$where the $\oplus$ on the L.H.S. is a direct sum of C*-algebras (derived from the notion of direct sum of algebras), and the $\times$ on the R.H.S. is a direct product of groups.
We also know that $K_0$ is a split-exact functor, in the sense that if we have a split short exact sequence of C*-algebras: $$ 0\longrightarrow I\longrightarrow A\longrightarrow B \longrightarrow 0$$ then we get a split short exact sequence of Abelian groups: $$ 0\longrightarrow K_0\left(I\right)\longrightarrow K_0\left(A\right)\longrightarrow K_0\left(B\right) \longrightarrow 0$$
Now take for example the unitzation $\tilde{A}$ of any C*-algebra $A$ (unital or not). We know that $$\tilde{A} \neq A\oplus \mathbb{C}$$ as algebras (unless $A$ is unital) (so we cannot use equation $(\ast)$), but that at the level of vector spaces this is true. We also know using the above fact that $K_0$ is split-exact that $$K_0(\tilde{A}) = K_0(A)\times \mathbb{Z}$$ which would have been the result if we could have applied equation $(\ast)$.
Thus arises the following question: If we have a C*-star algebra $C$ such that at the level of vector spaces we have $$ C = A\oplus B$$ for some two other C*-algebras $A$ and $B$ but such that at the level of algebras $$C \neq A \oplus B$$ then could we still apply equation $(\ast)$ to conclude $$ K_0(C) = K_0(A) \times K_0(B)$$?
If yes, would the proof be the same as the one which proves $(\ast)$? if not, could anyone provide a counter-example?
Consider $M_2(\mathbb C)$. As vectorspace it is isomorphic to $\mathbb C ^4$. But $K_0(M_2(\mathbb C)) = K_0(\mathbb C) = \mathbb Z \neq \mathbb Z^4 = K_0(\mathbb C)^4.$