Thanks a lot! +1. In the contravariant case, does it suffice to replace $kX$ with its dual?

@lcv - Thank you for your interest! Well, $f$ is the Lagrange interpolation polynomial iff $A$ is diagonalizable, iff the roots of $\mu$ are simple. For instance, if $A^2=0$ but $A\ne0$, then $\mu(x)=x^2,f(x)=1+x,e^A=I+A$, but Lagrange's polynomial is $g(x)=1$, which gives $g(A)=I\ne e^A$.

@lcv - No, I don't know. All I can say is this: If $A$ is a complex square matrix and $\mu$ is its minimal polynomial, then there is a unique univariate polynomial $f$ with complex coefficients such that $\deg f<\deg\mu$ and $e^A=f(A)$. Moreover there is a simple expression for $f$ in term of the roots of $\mu$ and their multiplicities.

@DenisNardin - Here is my (perhaps incorrect) argument for the claim "yes to Question 1 would imply yes to Question 2": Let $C\in\mathsf{CRing}$ be arbitrary, write $C=\lim_iC_i$ with $C_i$ noetherian, set $\text{Hom}:=\text{Hom}_{\mathsf{CRing}}$, and note $$\text{Hom}(B,C)\simeq\lim_i\text{Hom}(B,C_i)\simeq\lim_i\text{Hom}(A,C_i)\simeq\text{Hom}(A,C).$$ This implies that $f$ is an isomorphism. Thanks for telling me what's wrong with this, and for spelling out your argument.

@YCor - You mean that "limit" and "projective limit" are synonymous, and so are "colimit" and "inductive limit", not the other way around. Don't you?

Thanks for your answer and your comment! If I'm not mistaken your post also gives a negative answer to Question 1. Don't you think so? [In your answer I think $x=a$.]

@YCor - Thanks! The question is about limits, not colimits. In question 1 I wrote "Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?". Do you think it's not a clear enough definition of what is meant by "limit"?

Thank you very much for this wonderful answer! (Minor typo: "if $\mathfrak{m}$ IF finitely generated" should be "if $\mathfrak{m}$ IS finitely generated".)