Consider the matrix $$ A=\left(\begin{matrix}a&b\\0&a\end{matrix}\right),\qquad A^n=\left(\begin{matrix}a^n&b(na^{n-1})\\0&a^n\end{matrix}\right) $$ The exponential series is $$ \exp(X)=\sum_{k=0}^\infty \frac{1}{k!}X^k $$
Thus take the formal exponential of $A$ I get $$ \exp(A)=\sum_{k=0}^\infty \frac{1}{k!}\left(\begin{matrix}a^k&b(ka^{k-1})\\0&a^k\end{matrix}\right) $$ Since $ka^{k-1}$ is just the derivative of $a^k$, and $\frac{d}{dx}e^x=e^x$, I get $$\tag{*} \exp(A)=\left(\begin{matrix}e^a&be^a\\0&e^a\end{matrix}\right) =e^a\left(\begin{matrix}1&b\\0&1\end{matrix}\right) $$
An easier computation that gives some consistency would be $$ \begin{align} \exp\left(\begin{matrix}a&b\\0&a\end{matrix}\right) &=\exp\left[\left(\begin{matrix}a&0\\0&a\end{matrix}\right) +\left(\begin{matrix}0&b\\0&0\end{matrix}\right)\right]\\ &=\exp\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\exp \left(\begin{matrix}0&b\\0&0\end{matrix}\right) =\left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right) \left(\begin{matrix}1&b\\0&1\end{matrix}\right) \end{align} $$ since in exponential expansion of the second factor only the first two terms are relevant.
(Q1) Makes this computation any sense? Is it generalisable for the Jordan normal form?
(Q2) If we look at $$ A=\left(\begin{matrix}a&b\\0&a\end{matrix}\right) =\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right) $$ makes the following any sense, or can we extract anything interesting from it? $$ e^a\left(\begin{matrix}1&b\\0&1\end{matrix}\right) =\exp A = \left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right)^{\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right)} $$
(Q1), yes this makes sense, since $\left(\begin{matrix}a&0\\0&a\end{matrix}\right)$ and $\left(\begin{matrix}0&b\\0&0\end{matrix}\right)$ commute.
(Q2), $\left(\begin{matrix}e^a&0\\0&e^a\end{matrix}\right)^{\left(\begin{matrix}1&b/a\\0&1\end{matrix}\right)}$ makes no sense.