Let $R$ be a (unital) ring with the invariant base number property, and let $K$ be a left $R$-module such that $K\oplus R^{n}\cong R^n$ for some $n\geq 0$. Is it true that $K=0$?
I know that this is true if for example $R$ is commutative, because then we can tensor with $R_{\mathfrak{m}}/\mathfrak{m}R_{\mathfrak{m}}$ for a maximal ideal $\mathfrak{m}$ and reduce it to vector spaces. Does it still hold if we only assume the IBN-property?
Not in general, even for $n=1$.
Let $R=\mathbb{Z}\langle x,y\mid yx=1\rangle$, the ring generated by two noncommuting variables $x$ and $y$ subject to the single relation $yx=1$. This ring has a basis $\{x^iy^j\mid 0\leq i,j<\infty\}$ over $\mathbb{Z}$ with multiplication of basis elements easily deducible from the relation (just cancel as many instances of $yx$ as possible).
There is a surjective ring homomorphism $\varphi:R\to\mathbb{Z}$ with $\varphi(x)=\varphi(y)=1$ and so $R$ has the IBN property (a ring has IBN if it has a homomorphic image with IBN).
Right multiplication by $x$ is a left $R$-module homomorphism $\alpha:~_RR\to~_RR$ and is surjective since it has a right inverse given by right multiplication by $y$. Since $_RR$ is projective, $\alpha$ splits and so $$_RR~\cong~_RR\oplus\ker(\alpha).$$
But $\ker(\alpha)\neq0$, since $(xy-1)x=0$, so $xy-1\in\ker(\alpha)$.