Since the function $\exp(-t)/t$ is monotone decreasing for $t>0$ it is immediate to get $$f(x) := \int_x^{2x} \frac{\exp(-t)}{t} dt \le \exp(-x)$$ for $x>0$. Using $$E_1(x) = -\gamma - \log(x) - \sum_{k=1}^{\infty} \frac{(-x)^k}{k\; k!}$$ for $x>0$ (see Wikipedia for that formular) and realizing $$f(x)=E_1(x)-E_1(2x)$$ we even get $$\lim_{x \to 0} f(x) = \log(2).$$
Question:
Is there a $C \in (0,\log(2)]$ such that $$C \exp(-x) \le f(x) \le \exp(-x)$$ for all $x>0$?
For $x>0$, let $F(x) := e^x f(x)$. Then $$ F'(x) = e^x \left[ {f(x) + \frac{{e^{ - 2x} }}{{2x}} - \frac{{e^{ - x} }}{x}} \right] = - e^x \int_x^{2x} {\frac{{e^{ - t} }}{{t^2 }}dt} < 0 $$ where the second equality follows by integration by parts. Thus $$ 0 = \mathop {\lim }\limits_{t \to + \infty } (e^t f(t)) < e^x f(x) < \mathop {\lim }\limits_{t \to 0 + } (e^t f(t)) = \log 2 $$ for any $x>0$. The limit at infinity follows from http://dlmf.nist.gov/6.12.E1 and the definition of $f(x)$.
Addendum: By a simple change of integration variables ($t=x(s+1)$) $$ e^x f(x) = \int_0^1 {\frac{{e^{ - xs} }}{{s + 1}}ds} , $$ which leads to the same claim (use dominated convergence).