If $\lim_{x\to\infty}\frac{f'(x)}{f(x)}=1$, is $f$ asymptotic to $\exp$?

No. If we set $f(x) = e^{g(x)}$, then $\frac{f'(x)}{f(x)} = g'(x)$, and so the question is asking whether $\lim_{x\to\infty} g'(x) = 1$ implies that $g(x)$ is linear. But there are plenty of functions $h(x)$ such that $\lim_{x\to\infty} h(x)=1$ yet $h(x)$ is not identically $1$, and we can simply let $g(x)$ be an antiderivative of any such $h(x)$. For example, setting $h(x) = 1+\frac1x$ yields $g(x) = x+\ln x$ and so $f(x) = xe^x$, as commented by David Mitra.

Let $f(x)=xe^x$. Then $$\frac{f'(x)}{f(x)}=\frac{xe^x+e^x}{xe^x}=1+\frac{1}{x}\xrightarrow{x\to\infty}1,$$ but $f$ is not asymptotic to an exponential function.

The problem is exactly what you state; $C_1$ and $C_2$ depend on $\epsilon$.