Note that we have $$\frac{b^h-1}{h}=\frac{e^{h\log(b)}-1}{h} \tag 1$$
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality
$$e^x\ge 1+x \tag 2$$
From $(2)$ (along with the property $e^xe^{-x}=1$) it is easy to see that for $x<1$
$$e^x\le \frac{1}{1-x} \tag 3$$
Using $(2)$ and $(3)$, we can bound $(1)$ as
$$\log(b) \le \frac{e^{h\log(b)}-1}{h}\le \frac{\log(b)}{1-h\log(b)}$$
whereupon application of the squeeze theorem yields the coveted limit
$$\lim_{h\to 0}\frac{b^h-1}{h}=\log(b)$$
And we are done!
This might not exactly be the type of answer you are looking for, but I think it's accurate so I'm writing it here anyway. Often in math, we can encounter objects with multiple (provably equivalent) definitions, some more natural than others, and some can be better in giving an intuitive insight into how the object in question works. Once we've chosen a definition, the others can then be proven as theorems.
If I guess correctly, your definition of the exponential function $\exp$ is given by first defining
$$\log(x)=\lim_{n\to\infty}n(x^{1/n}-1),$$
and then defining $\exp$ to be the inverse function of the logarithm. But this is not a very enlightening definition. A much more common definition would literally be that $\exp$ is the unique function $f:\mathbb R\to\mathbb R$ satisfying $f'(x)=f(x)$ for all $x$, and $f(0)=1$. Proving such a function exists and is unique takes some work, then showing that this is indeed equivalent to your definition takes some more. But once this is done, we can then accept, as a fact, that $\exp$ is the function that is meant to satisfy $\exp'=\exp$. We can define this in multiple ways, but this is the core property that is central to the exponential function---some might say, the defining property of the exponential function.
So your question is:
Are there easier ways of proving that $\dfrac{d}{dx}e^x=e^x$?
I would say yes: by assuming it is the case, by taking it to be true by definition. Showing that other definitions are equivalent to this definition is nontrivial and needs to be done of course, but as you observed might not necessarily be very intuitive or give much insight. To understand what the exponential function is deeply, this definition is the way to go.
To show that the definition $\frac{d}{dx}e^x=e^x$ is equivalent to the "arithmetic definition" $e^{a+b}=e^ae^b$ is an interesting problem. To go from the former to the latter, see the link provided by KM101 in the comments. On the other hand, let's try to start from the latter and try to go to the former. Now, differentiation deals with the "local growth rate" of a function with response to small changes in $x$, and we have an additive definition, so we consider
$$\exp(x+\Delta x)=\exp(x)\exp(\Delta x)$$
for a small $\Delta x>0$. Now in the definition of the derivative, we consider
$$\exp(x)\lim_{\Delta x\to 0}\left(\frac{\exp(\Delta x)-1}{\Delta x}\right).$$
To show that this is indeed $\exp(x)$, all we need to do is to show that the latter limit is $1$. Note that the limit is actually just $\exp'(0)$. Consider what we've done so far: we've reduced the proof of $\exp'=\exp$, which is information regarding the "growth rate" of $\exp$ globally, into the proof of just $\exp'(0)=1$, at a single point!
So how can we prove this, knowing just $\exp(a+b)=\exp(a)\exp(b)$? Well... unfortunately we can't. See, if we define $f'(0)=k$ for any arbitrary real number $k$, where $f$ satisfies $f(a+b)=f(a)f(b)$, we will get a perfectly well-defined function $f$ (you can try to show this). But when we make the choice that $f'(0)=1$ (or in other words $k=1$), then we end up with a function which is its own derivative. Indeed, this is the property that motivates the choice $f'(0)=1$.
In hindsight, if $f'(0)=k$ and $f(a+b)=f(a)f(b)$ for all $a,b$, we can describe $f$ in general: it is simply $f(x)=e^{kx}$.
Best Answer
As I can see from your edit, you noticed that the only difficult part is to show that $\exp'(0) = 1$, that is, $\lim_{h\to 0}\left({\dfrac{\exp(h)-1}{h}}\right) = 1$.
So here's my proof, using only the definition of the exponential function and elementary properties of limits.
We use the following definition of the exponential function:
\begin{align*} &\exp : \mathbb{R} \to \mathbb{R}\\ &\exp(x) = \lim_{k \to +\infty} \left(1 + \frac{x}{k}\right)^k \end{align*}
Let's define \begin{align*} &A : \mathbb{R}^* \to \mathbb{R}\\ &A(h) = \frac{\exp(h) - 1}{h} - 1 \end{align*}
We're going to show that $\lim_{h \to 0} A(h) = 0$. This will imply that $\lim_{h \to 0} \frac{\exp(h) - 1}{h} = 1$ and consequently, that $\exp'(0) = 1$.
Let's show that for all $h \in [-1,1]\setminus\{0\}$, $|A(h)| \leq |h|$
Let $h \in [-1,1]\setminus\{0\}$. We define the sequence $(u_k)_{k \in \mathbb{N}^*}$ by \begin{align*} u_k = \frac{\left(1+\frac{h}{k}\right)^k - 1}{h} - 1 \end{align*}
From the definition of the exponential function and from the rules of addition and multiplication of limits, we get:
\begin{align*} \lim_{k \to + \infty} u_k = A(h) \end{align*}
The continuity of the absolute value function then brings:
\begin{align*} \lim_{k \to + \infty} |u_k| = |\lim_{k \to + \infty} u_k| = |A(h)| \end{align*}
If we manage to show that after a certain rank, $|u_k| \leq |h|$, we'll be able to conclude that $|A(h)| = \lim_{k \to +\infty}|u_k| \leq |h|$.
For $k \in \mathbb{N}^*$, we have
$$ u_k = \frac{\left(\sum\limits_{i=0}^{k} \binom{k}{i} \left(\frac{h}{k}\right)^i \right) - 1 - h}{h} = \frac{1}{h}\sum_{i=2}^k \frac{\binom{k}{i}}{k^i} h^i = h \sum_{i=2}^k \frac{\binom{k}{i}}{k^i} h^{i-2} $$
The triangle inequality brings: $$ |u_k| \leq |h| \sum_{i=2}^k \frac{\binom{k}{i}}{k^i} |h|^{i-2} $$
We have $h \in [-1,1]$. So $|h|^{i-2} \leq 1$ for every $i \in \mathbb{N}$ such as $i \geq 2$. Moreover, for $k,i \in \mathbb{N}\setminus\{0,1\}$: $$ \frac{\binom{k}{i}}{k^i} = \frac{\prod\limits_{j=0}^{i-1}(k-j)}{i!\prod\limits_{j=0}^{i-1}k} \leq \frac{1}{i!} \leq \frac{1}{2^{i-1}} $$ Therefore, as soon as $k \geq 2$: $$ |u_k| \leq |h| \sum_{i=2}^k \frac{1}{2^{i-1}} = |h| \sum_{i=1}^{k-1} \frac{1}{2^i} = |h| \left(\frac{1-\frac{1}{2^k}}{1-\frac{1}{2}} - 1\right) = |h| \left(1-\frac{1}{2^{k-1}}\right) \leq |h| $$
Hence: $$ |A(h)| = \lim_{k \to \infty} |u_k| \leq |h| $$ This is true for all $h \in [-1,1]\setminus\{0\}$. Therefore: $$ \lim_{h \to 0} A(h) = 0 $$ which shows that $\exp'(0) = 1$.