Let $X$ and $Y$ be different groups. Let $G$ be a group acting on $X$. Is that possible that $G$ act on $X \times Y$ by $g \cdot (x,y) = (g \cdot x,y)$?
I know it sounds silly and unnatural, but for me, it seems like not violating the definition of group action. Could someone explain it to me?
Yes, this is a valid group action. I assume you mean "$X$ and $Y$ be different sets (instead of groups)". You may think of $X\times Y$ as $|Y|$ disjoint copies of $X$, and $G$ acts on it by acting on each individual copy.
For example, $S_3$ naturally acts on an equilateral triangle $\Delta$, as well as two disjoint equilateral triangles $\Delta_1\sqcup \Delta_2 \simeq \Delta\times \{1, 2\}$, etc.
Such actions enjoy the special property that if $(x,y)$ and $(x', y')$ are in the same orbits, then $y=y'$. Hence the $Y$-coordinate helps classify the orbits of the group action.
Response to the comment about ring actions:
I'm not sure what you are looking for. Group action and ring action are just different... if I have to say, given two groups $G, H$ acting on $X, Y$ respectively, there is a group action of $G\times H$ on $X\times Y$. Similarly, given two rings $R, S$ acting on abelian groups $M, N$ respectively, there is a ring action of $R\times S$ on $M\times N$. Here might be the key difference when $H$ (or $S$) is missing: The "trivial" group is $H=\{1\}$, and we have $G\times H\simeq G$, while the "trivial" ring is $S=\mathbb Z$ (not the zero ring in this case, as it has no action on general $N$), and we have $R\times \mathbb Z\not\simeq R$. Similarly, you do have $R\otimes S$ acting on $M\otimes N$, and in this case $R\otimes\mathbb Z\simeq R$, hence you don't need any special $S$ for tensor product.
In other words, the initial object in the category of groups is an identity for Cartesian product, while the initial object in the category of rings is not.