I'm reading Rordam's book "An Introduction to K-theory for c*-algebras".
Given two C*-algebras $A,B$, the book proves that $K_0(A\bigoplus B)\cong K_0(A)\bigoplus K_0(B)$.
I'm assuming that the book means $A\bigoplus B$ to be a direct sum over C*-algebras (as opposed to vector spaces).
Now to calculate $K_0(C_0(\mathbb{R}))$. The book has a "warm up" exercise that shows for $X$ Hausdorff with separation $X=X_1\cup X_2$, then $C_0(X)\cong C_0(X_1)\bigoplus C_0(X_2)$ (as both vector spaces and c*-algebras). I'm then supposed to use that result and the knowledge that $K_0(C(S^1))\cong\mathbb{Z}$ to conclude that $K_0(C_0(\mathbb{R}))=0$.
Aside from $S^1$ being connected so the warm up is a bit misleading, my issue is that I can show that $C(S^1)\cong C_0(\mathbb{R})\bigoplus\mathbb{C}$ as vector spaces, but the direct sum does not hold as algebras (in partciular $C_0(\mathbb{R})$ is not unital so its unitization is not isomorphic to itself plus a copy of $\mathbb{C}$).
I then am not sure how one is supposed to conclude that $K_0(C_0(\mathbb{R}))\cong 0$ or where I am misunderstanding something.
The warm up is not misleading, because after you compute $K_0(C_0(\Bbb{R}))$ you need to compute $K_0(C_{0}(U))$ and you'll need this result for the second computation.
As you said, $C(S^1)$ and $C_0(\Bbb{R})\oplus \Bbb{C}$ are not isomorphic as $C^*$-algebras.
However, consider the short exact sequence:
$0 \to C_0(\Bbb{R}) \to C(S^1) \to \Bbb{C} \to 0$, you know it is split exact just because it is the split exact sequence related to the unitalization of a $C^*$ algebra.
You also have this argument as an example in the book, see Example 4.3.5.
Now you can conclude that $\Bbb{Z} \cong K_0(C(S^1)) \cong K_0(C_0(\Bbb {R}))\oplus \Bbb{Z}$ (because $K_0$ is split exact functor).
As $K_0(C_0(\Bbb{R}))$ is an abelian sub-group of $\Bbb{Z} \cong K_0(C(S^1))$, it must be zero.
I hope this helps.