When I look at different proofs of the Jordan-Chevalley decomposition of a matrix, the minimal hypothesis I usually found is about the perfection of the field over which such decomposition occurs (e.g. Wikipedia article). But it seems to me that the gist of the proofs is not about the field but about the type of characteristic polynomial of the matrix.
More precisely, by using the two following definitions for a general ring $A$
Definition 1 A $n\times n$-matrix $M$ with coefficients in the ring A is semisimple if $A^n$ is a semi-simple $A[M]$-module (here I am using the definition provided in the answer of @ಠ_ಠ below)
Definition 2 A separable polynomial $P$ of $A[X]$ is a polynomial whose discriminant is invertible in $A$
Then is the following statement true?
A matrix $M$ over $A$ have a unique Jordan-Chevalley decomposition if its characteristic polynomial divides a power of a separable polynomial.
Edit By Jordan decomposition I mean here the additive one, i.e. there exist $M_s$ and $M_n$, two $n\times n$ $A$-matrices respectively semi-simple and nilpotent, such that $$M=M_s+M_n$$
If it is not, what kind of general property should the ring $A$ have to make it true (like $A$ is a domain, etc.)
Edit 2 I believe my definition 1 is equivalent in a domain to the fact that the matrix is diagonalizable in an algebraic extension of the fraction field which is equivalent to the fact that the matrix is cancelled by a separable polynomial (definition 2) in $A[X]$
If $K$ is a commutative ring and $V$ is a $K$-module, then a choice of endomorphism $f \in \text{End}_K(V)$ is the same as a representation of the $K$-algebra $K[f]$ on $V$. This is proved in Aluffi's Algebra text in section VI.7.1. Then we say $f$ is simple (resp. semisimple) if $V$ is simple (resp. semisimple) as a $K[f]$-module.
Sorry I can't help with the rest of your questions.