For smooth varieties $X$ and $Y$ over a field $k$, it is pretty rare to have that the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism (or surjective).

For instance, suppose $X$ is smooth and proper over $k$, and the product map above is surjective for $Y=X$. Consider the class of the diagonal $\Delta_X$ in $K_0(X\times_kX)$ (this makes sense because $K_0$ is the same for vector bundles and coherent sheaves). Then the class of $\Delta_X$ is expressible as a finite sum $\sum_ia_i\otimes b_i$ with $a_i,b_i\in K_0(X)$.

View elements $\alpha\in K_0(X\times_kX)$ as correspondences, i.e., as endomorphisms of $K_0(X)$, given by $x\mapsto (p_2)_*(p_1^*(x)\cdot\alpha)$, where $\cdot$ is the multiplication in $K_0$ (the direct image $(p_2)_*$ exists because $X$ is proper). The class of the diagonal gives the identity endomorphism, while the class of an element which is the image of $a\otimes b$ is rather special: it is of the form $x\mapsto \chi(x\cdot a)b$, where $\chi:K_0(X)\to {\mathbb Z}$ is the Euler characteristic ($\chi$ is the direct image map $K_0(X)\to K_0(k)$). Using this, one can deduce that $K_0(X)$ must be free abelian of finite rank, and $a_i$ (and also $b_i$) form a $\mathbb Z$-basis (and in fact, for the non-degenerate pairing $(x,y)\mapsto \chi(x\cdot y)$, they form dual bases). This argument, in some form, is well-known in some circles; this ``dual bases'' formulation appears in work of Ivan Panin.

Thus it is easy to give examples of such $X$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is not surjective, e.g., a smooth projective curve of positive genus over an algebraically closed field, or (more tricky) a complex Enriques surface ($K_0$ is finitely generated, but has torsion).

I do not know a counterexample to the following:

let $X$ be a smooth (say, proper) variety over a field $k$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is an isomorphism; then for any smooth variety $Y$, the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism.