Is $e^x$ the only non-trivial function for which the differential operator is the identity operator? Given that the derivative of a function is defined as:
$$\frac{d}{dx}f(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\tag{1}\label{1}$$
and:
$$\frac{d}{dx}e^x=e^x\tag{2}\label{2}$$
Can we get $e^x=f(x)$ from the following condition?
$$f(x)=\frac{d}{dx}f(x)=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\tag{3}\label{3}$$
Could the differential operator be interpreted as the identity operator for $f(x)=e^x$? 

Is $e^x$ the only non-trivial function for which the differential operator is the identity operator?
 A: Using elementary methods for "solving a differential equation" we can find all such functions.  
Suppose $f(x)$ is a differentiable function $\mathbb R \to \mathbb R$ such that $$f'(x) = f(x)\quad\text{for all }x.\tag{1}$$
Consider the function $g(x) = e^{-x}f(x)$.  Use the product rule to compute
\begin{align}
g'(x) &= \frac{d}{dx}\big(e^{-x}f(x)\big) = f(x)\frac{d}{dx}\big(e^{-x}\big)+e^{-x}\frac{d}{dx}\big(f(x)\big)\\ &=
-f(x)e^{-x} + e^{-x} f(x) = 0
\end{align}
But if $g'(x) = 0$ for all $x$, it follows that $g(x) = c$ for some constant $c$.  (Proof from the mean value theorem.)  Therefore
\begin{align}
g(x) &= c
\\
e^{-x}f(x) &= c
\\
f(x) &= c e^{x}
\end{align}
The functions $ce^{x}$ are the only solutions for $(1)$.  There are of course infintely many solutions, one for each constant $c$.  

The same method shows that the solutions for functions $f : \mathbb C \to \mathbb C$ that satisfy $f'(z)=f(z)$ for all $z \in \mathbb C$ are the functions $f(z) = c e^{z}$.  One for each complex constant $c$.
A: Operators aren't equal somewhere but unequal somewhere else; that's sloppy terminology. What we'd usually say is functions of the form $ke^x$ are the only eigenfunctions of $\frac{d}{dx}$ with eigenvalue $1$, or that they're the only functions on which $\frac{d}{dx}$ has the same action as the identity operator does, or they're its only fixed points, or they're the only elements of the kernel of $\frac{d}{dx}-I$.
Actually, I think that talk of "action" will probably raise eyebrows too; I should say these are the functions with the same image under both operators.
A: If consider functions of the form $ke^x$, then YES.
Assume $f(x)>0$. As $$\mathrm{\frac{d(\log(f(x)))}{dx}}=\frac{1}{f(x)}\cdot\mathrm{\frac{df(x)}{dx}}=1,$$
We have $$\int{\mathrm{\frac{d(\log(f(x)))}{dx}}}\mathrm dx=\int 1\mathrm dx=x+C,$$
where $C\in \mathbb{R}$ is a constant number. Now The left hand side is $\log(f(x))$ and the right hand side is $x+C$. Thus
$$f(x)=e^{\log(f(x)}=e^{x+C}=e^C\cdot e^x=ke^x,$$
If $f(x)<0$, then consider $\mathrm{\frac{d(\log(-f(x)))}{dx}}$. In a similar way, we have 
$$f(x)=-e^C\cdot e^x=ke^x,$$
where $k=-e^C<0$.
where $k=e^C\ge 0$ is a constant number too.
