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 $\exp'(0) = 1$.
