Finitely generated $K_0$ of $C^*$-algebras Let $A$ and $B$ be $C^*$-algebras. Let $A_0$ be a dense *-subalgebra of $A$, and let $B_0$ a dense *-subalgebra of $B$. Assume that $A_0$ is isomorphic to $B_0$. 
Finally, assume that $K_0(A)$ is finitely generated. Must $K_0(B)$ also be finitely generated? 
The isomorphism between $A_0$ and $B_0$ is an algebraic isomorphism between algebras over the field of complex numbers; it need not be a $*$-isomorphism. 
 A: No. To obtain a counterexample, you just need a C*-algebra $A$ with finitely generated K-theory and a quotient $A/I$ of $A$ which does not have finitely generated K-theory and the quotient algebraically preserves some dense subalgebra of $A$. Here is the easiest example of this situation that I could think of. 
Let $A=C[0,1]$ and let $B$ be the C*-algebra of convergent sequences. Then $K_0(A)$ is finitely generated while $K_0(B)$ is not. Define the surjective map $\pi:A\rightarrow B$ by $\pi(f)(n)=f(1/n).$  Let $A_0\subseteq A$ be polynomials and let $B_0=\pi(A_0),$ which is a dense subalgebra of $B.$  Since $\pi$ is injective on $A_0$ it defines a *-isomorphism between $A_0$ and $B_0.$  
A: We know that $K_0(A) \neq K_0(B)$ in general. Take for example, what Handelman pointed out, the full and reduced group $C^*$-algebras of a group $G$ which is not $K$-amenable, i.e. $K_0(C^* G) \neq K_0(C^*_r G)$.
It is fair to assume that we may find an example where $K_0(A)=0$ but $K_0(B) \neq 0$.
If you accept this, then you immediately have a counterexample.
Take $a_0:=\bigoplus_{\mathbb{N}} A_0$ with two distinct closures
$a:=\bigoplus A$ ($C^*$-direct inifinite sum of $A$) and similarly $b:=\bigoplus B$.
Then $a_0$ is dense in $a$ and $b$.
But $K_0(a) = \bigoplus K_0(A) =0$ is finitely generated, and $K_0(b)= \bigoplus K_0(B)$ is infinitely generated.
That is why I am confident that your question is answered to the negative.
