If $f = a \exp$, then from the results
we know that conditions (1) and (2) hold.
Conversely, assume that the conditions (1) and (2) hold. Using the results
and the assumption $f' = f$, we have
\begin{equation}
\frac{d}{d x} f(x) \exp(- x)
= f'(x) \exp(- x) - f(x) \exp(- x)
= f'(x) \exp(- x) - f'(x) \exp(- x)
= 0
\end{equation}
for every $x \in \mathbb{R}$. That is, the function
\begin{equation}
x \mapsto f(x) \exp(- x)
= \frac{f(x)}{\exp(x)}
\end{equation}
is constant everywhere on $\mathbb{R}$. Since $f(0) = a$, then $f(0) \exp(-0) = a \exp(0) = a$. That is, $f(x) \exp(-x) = a$ and equivalently $f(x) = a \exp(x)$ for every $x \in \mathbb{R}$. $\square$