is there any example, s.t. $K=\mathbb{C}$ is a field and we have two non-isomoprhic finite groups $G$ and $H$, s.t. the group ring $K[G]$ is isomorphic to $K[H]$ as a $K$-algebra?
The idea of this question is that if I give you the group ring $K[G]$, can you determine the group $G$?
Here Are groups completely determined by their representations? you can see that there exists an example for the case $K=\mathbb{Z}$.
Maschke's theorem together with the Artin-Wedderburn theorem tells you that $\mathbb{C}[G]$ is isomorphic to a product of matrix algebras $M_{d_i}(\mathbb{C})$ where $d_i$ are the dimensions of the irreducible representations of $G$. So two group algebras of finite groups over $\mathbb{C}$ are isomorphic iff the dimensions of their complex irreducible representations are the same.
In particular this is true for any two finite abelian groups of the same cardinality, so the smallest example is $G = C_2 \times C_2, H = C_4$ (this is the smallest example of a pair of non-isomorphic groups of the same order). The smallest nonabelian example is $G = Q_8, H = D_4$, whose representations have dimensions $1, 1, 1, 1, 2$ (and which even famously have the same character table).