Suppose that $A$ is an invertible matrix and
$$x = \operatorname{argmin}_{x \in X}\lvert Ax - y\rvert^2,$$
then does it mean that $Ax = \operatorname{Proj}_X(y)$ like in the definition of projection? I mean, the definition gives that it would be the case if
$$Ax = \operatorname{argmin}_{Ax \in X}\lvert Ax - y\rvert^2.$$
But does the above relation imply this one?
No, this is definitely not correct. The solution is in fact $$x=A^{-1}\textrm{Proj}_{Y}(y)$$ where $Y=AX$; that is, $Y$ is the set $$Y\triangleq AX = \{ Ax \,|\, x\in X\} = \{ y \,|\, A^{-1}y\in X \}.$$ Whether or not an efficient projection can be constructed in this case, or whether that projection can utilize an existing $\textrm{Proj}_X$ implementation, will depend on the specific structure of $X$.