I want to compute the K-theory of $C_{b}[0,\infty)$, the algebra of bounded, uniformly continuous functions on $[0,\infty)$, by considering the exact sequence $0\rightarrow C_{b,0}(\bigcup_{n\geq 0}(2n-1,2n))\rightarrow C_{b}[0,\infty)\rightarrow C_{b}(\bigcup_{n\geq 0}[2n,2n+1])\rightarrow 0$, where the maps are inclusion and restriction.
It boils down to looking at the homomorphism $K_{0}(C_{b}(\bigcup_{n\geq 0}[2n,2n+1]))\rightarrow K_{1}(C_{b,0}(\bigcup_{n\geq 0}(2n-1,2n)))$ in the K-theory six term exact sequence. This is a homomorphism $\prod\mathbb{Z}\rightarrow\prod\mathbb{Z}$ but I am stuck on showing that it is surjective and has kernel $\mathbb{Z}$.
In the title you write "bounded continuous functions", but in the question itself you write "bounded, uniformly continuous functions". Since I do not know which of these you actually really mean, I will give answers to both versions.
Let us denote your exact sequence by $0 \to J \to A \to A/J \to 0$ as it is usually done.
Firstly, let us consider its uniformly continuous version. As you have probably correctly noticed, we do have $K_0(J) = 0$ and $K_1(A/J) = 0$. But you missed a crucial point for the other two groups $K_1(J)$ and $K_0(A/J)$. They are both isomorphic to the group of all bounded, integer-valued sequences $\ell^\infty_\mathbb{Z}$. But you wrote $\prod \mathbb{Z}$ which does not encompass the boundedness. Now you conjecture that the boundary homomorphism $\ell^\infty_\mathbb{Z} \to \ell^\infty_\mathbb{Z}$ of that 6-term sequence is surjective with kernel $\mathbb{Z}$, which would give us $K_0(A) \cong \mathbb{Z}$ and $K_1(A) \cong 0$. Though you are correct with the kernel of the boundary homomorphism being $\mathbb{Z}$ and therefore $K_0(A) \cong \mathbb{Z}$, you are completely wrong with the surjectivity. In fact, the group $K_1(A)$ is highly non-trivial.
As a remark: what is actually the K-theory of the $C^\ast$-algebra of all bounded, uniformly continuous functions? Let me answer this question if the space is a Riemannian manifold $M$ of bounded geometry (meaning that its injectivity radius is uniformly positive and the curvature tensor and all its covariant derivatives are bounded in norm). Then we may identify $K_0(C_u(M))$ with the following: every element of it is of the form $[E] - [F]$, where $E, F \to M$ are complex vector bundles of bounded geometry over $M$ (meaning that they are equipped with a hermitian metric and compatible connection such that their curvature tensor and all its covariant derivatives are bounded in norm) and $[\cdot]$ denotes $C_b^\infty$-isomorphism classes, i.e., using isomorphisms $\varphi\colon E \to E^\prime$ that are bounded, their inverses are bounded, and such that $\nabla^E - \varphi^\ast \nabla^{E^\prime}$ is a bounded tensor and also all its covariant derivatives are bounded. One can give a similar interpretation of $K_1(C_u(M))$ using vector bundles over $S^1 \times M$ which are trivial in a neighbourhood around $1 \in S^1$.
Using the above you may now interpret $K_0(A)$ and $K_1(A)$ (though $[0,\infty)$ is not a manifold of bounded geometry due to its boundary, we should probably still get the same interpretation here). By the way, the above mentioned interpretation is actually part of my forthcoming thesis.
Now let us consider the other case where you really mean the bounded continuous functions (i.e., not only uniformly continuous functions). But the problem here is that I can't say that much since I can't even compute the $K$-theory of $A/J$ in this case. It is tempting to say that we may homotope each interval to a point and then again you get, e.g., $K_1(A/J) = 0$. But the problem here is that this homotopy of all intervals to a bunch of points is not an allowed homotopy for the $C^\ast$-algebra of all bounded continuous functions! (But it is one for the $C^\ast$-algebras of all bounded, uniformly continuous functions, i.e., this is the way how I computed the above groups in that case.) So I think that probably no one knows the $K$-groups in this case.