Why is $f(t) = e^{ta}$ differentiable in a unital Banach algebra?

Here is an elementary proof inspired on the classical proof for power series.

For $t \in \Bbb{R}$, put $$g(t):= \sum_{k=1}^\infty k\frac{t^{k-1}a^{k}}{k!}$$ $$S_n(t) := \sum_{k=0}^n \frac{t^ka^k}{k!}$$ $$R_n(t) := \sum_{k=n+1}^\infty \frac{t^ka^k}{k!}$$

All these series converge since $A$ is a Banach space.

Fix $t \in \Bbb{R}$ and let $\epsilon > 0$.

Note first that $\lim_n S_n'(t) = g(t)$, so there is $N_1$ such that $$n \geq N_1 \implies \Vert S_n'(t)-g(t)\Vert < \epsilon/3$$

Also, choose $N_2$ such that $$n \geq N_2 \implies \sum_{k=n+1}^\infty\frac{\Vert a \Vert^k}{k!} k (|t|+1)^{k-1} < \epsilon/3$$

Put $n:= \max \{N_1, N_2\}$. Choose $\delta> 0$ such that $$0 < |s-t| < \delta \implies \left\Vert \frac{S_n(s)-S_n(t)}{s-t}- S_n'(t)\right\Vert< \epsilon/3$$

Then for any $s \neq t$ with $|s-t| < \delta \land 1$, we have $$\left \Vert\frac{f(s)-f(t)}{s-t}- g(t)\right\Vert$$ $$\leq \left\Vert\frac{S_n(s)-S_n(t)}{s-t}-S_n'(t)\right\Vert+\Vert S_n'(t)-g(t)\Vert + \frac{\Vert R_n(s)-R_n(t)\Vert}{|s-t|}$$

But $$\left|\frac{s^k-t^k}{s-t}\right|= |t^{k-1}+ t^{k-2}s + \dots + ts^{k-2} + s^{k-1}| \leq k (|t|+1)^{k-1}$$ Hence $$\frac{\Vert R_n(s)-R_n(t)\Vert}{|s-t|}=\frac{\Vert \sum_{k=n+1}^\infty \frac{s^k-t^k}{k!} a^k\Vert}{|s-t|}\leq \sum_{k=n+1}^\infty \left|\frac{s^k-t^k}{s-t}\right|\Vert a\Vert^k/k! < \epsilon/3$$ and we conclude $$\left \Vert\frac{f(s)-f(t)}{s-t}- g(t)\right\Vert < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$

We thus have shown that $f'(t) = g(t) = a f(t)$ and the proof is done.

Reference: Conway's "Functions of one complex variable I" (I modified the proof I saw there).

Your intuition is right, we need functionals. This proof uses arguments like those used in the proof of the spectrum being non-empty in Banach algebras. Have a look:

Let $\tau\in A^*$. Then $\tau\circ f:\mathbb{R}\to\mathbb{C}$ is a continuous function and we have that $$\tau\circ f(t)=\tau(e^{ta})=\tau\bigg(\sum_{n=0}^\infty (ta)^n/n!\bigg)=\sum_{n=0}^\infty \frac{t^n\tau(a^n)}{n!}.$$ (we use continuity and linearity of $\tau$).

So $\tau\circ f$ is a power series and it converges everywhere, since all the above are well defined. As a power series, this is differentiable and we may differentiate term-by-term, so we have that $$\frac{d}{dt}(\tau\circ f)(t)=\sum_{n=1}^\infty\frac{t^{n-1}\tau(a^n)}{(n-1)!}=\sum_{n=0}^\infty\frac{t^n\tau(a^{n+1})}{n!} $$

Set $g(t)=af(t):\mathbb{R}\to A$. Note that for $\tau\in A^*$ it is $$\tau\circ g(t)=\tau\bigg(a\sum_{n=0}^\infty\frac{t^na^n}{n!}\bigg)=\tau\bigg(\sum_{n=0}^\infty\frac{t^na^{n+1}}{n!}\bigg)=\sum_{n=0}^\infty\frac{t^n\tau(a^{n+1})}{n!}$$ (we use continuity and linearity of $\tau$). Now observe that $$\frac{d}{dt}(\tau\circ f)(t)=\lim_{h\to0}\frac{\tau(f(t+h))-\tau(f(t))}{h}=\lim_{h\to0}\tau\bigg(\frac{f(t+h)-f(t)}{h}\bigg)=\tau(f'(t))$$ by continuity and linearity of $\tau$. By the above we get $\tau(f'(t))=\tau(g(t))$ for all $t\in\mathbb{R}$ and all $\tau\in A^*$. By Hahn-Banach we conclude that $f'(t)=g(t)$ for all $t$ and we are done.

We can also use some integration theory on Banach spaces. We have $$f'(t) = \lim_{t\to t_0} \frac{f(t)-f(t_0)}{t-t_0} = \lim_{t\to t_0} \sum_{n=0}^\infty \frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!}$$

Now, for every $t \in [t_0-1, t_0+1]$ by the mean value theorem we can dominate $$\left\|\frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!} \right\| \le \left|\frac{t^n-t_0^n}{t-t_0}\right| \frac{\|a\|^n}{n!} \le n(t_0+1)^{n-1} \frac{\|a\|^n}{n!} $$ which is an integrable function since $$\sum_{n=0}^\infty n(t_0+1)^{n-1} \frac{\|a\|^n}{n!} \le \|a\|\exp((t_0+1)\|a\|) < +\infty.$$

Therefore, by the Lebesgue Dominated Convergence Theorem we have $$f'(t) = \sum_{n=0}^\infty \lim_{t\to t_0}\frac{t^n-t_0^n}{t-t_0} \frac{a^n}{n!} = \sum_{n=0}^\infty nt_0^{n-1} \frac{a^n}{n!} = a\exp(t_0a) = af(t).$$