If $A = \mathrm{End}(V)$, where $V$ is an infinite-dimensional vector space over some field, then it's not hard to see that $A \cong A^2 \cong \dotsb$. In particular, the map $\mathbb{Z} \to K_0(A)$ sending $n$ to $n[A]$ has image zero.
Is it possible for the image of this map to be $\mathbb{Z}/n$ for other $n$? This would mean, for some $m$, that $A^m \cong A^{m+n}$, but that $A^{m'} \not\cong A^{m'+n'}$ for all $m'$ and all $n' < n$.
(Of course, it's a standard result that for $A$ commutative, $A^m \cong A^n$ means $m = n$, so $A$ is necessarily noncommutative here.)
There are very nice algebras called Leavitt algebras which furnish examples of non-IBN rings.
Given a non-IBN ring $R$, let $m$ be minimal such that $R^m\cong R^n$ for some $n>m$. Let $n$ also be minimal with respect to $m$. In this case $R$ is said to have "module type $(m,n)$.
Leavitt's existence theorem guarantees that there exists a Leavitt algebra of type $(m,n)$ for any prescribed $m<n$. If the slideshow is too sparse, check out the third chapter of this online book where Gene outlines everything.
I think this seems to be an example of what you are looking for, as $m'<m$ implies that $R^{m'}\ncong R^n$ for any $n$.