If $A \in M(d \times d, \mathbb{R})$ is invariant on subspace $E \subset \mathbb{R}^d$, does it follow that $e^{A}$ is also invariant on $E$?
I think so, since $e^{A} = \sum_{n=0}^{\infty}{A^n \over n!}$ and each $A^n$ is invariant on $E$ by induction. So the series is a sequence of partial sums of elements in $E$ and since $E$ is a subspace, i.e closed, the sequence converges also in $E$, meaning that $e^{A}(u) \in E$ for any $u \in E$, correct?
I am also not sure how to justify $$\Big(\sum_{n=0}^{\infty}{A^n \over n!}\Big)(u) = \sum_{n=0}^{\infty}{(A^n)(u) \over n!} $$
What you did is fine. Now, note that, since $E$ is a closed subset of $\mathbb R^d$ and, if $u\in E$, each $\sum_{n=0}^N\frac1{n!}A^n(u)$ belongs to $E$; then$$e^A(u)=\lim_{N\to\infty}\sum_{n=0}^N\frac1{n!}A^n(u)\in E.$$