If you are ok with $f$ being not continuous and not strictly decreasing, here's a simple enough counter-example. (With a bit more work you can make it continuous and strict, but it distracts from the point.)
Let $f(x)$ be the step-function defined by
$$ f(x) = \begin{cases} 2^{k} & x \in ( 2^{-k}, 2^{1-k}) \\ 2^{k} & x = 2^{-k}, \quad k \text{ is odd} \\ 2^{k+1} & x = 2^{-k}, \quad k\text{ is even} \end{cases} $$
You have that the integral $$ \int_a^{2a} f(t) ~dt = 1 $$ for any $a$.
You have that $2^{-k} f(2^{-k})$ alternating between 1 and 2 and does not converge.
Incidentally, the "comparability to $1/x$" in your question is superfluous. Since you assumed decreasing, you have that
$$ a f(2a) \leq \int_a^{2a} f(x) ~dx \leq a f(a) $$
which implies that
$$ \liminf xf(x) \geq \lim \int_a^{2a} f(x) ~dx $$
$$ \limsup xf(x) \leq 2 \lim\int_a^{2a} f(x)~ dx$$
so comparability is automatic, and also the existence of a sequence $x_n \to 0$ such that $x_n f(x_n)$ converges. (Of course $(2^{-n})$ need not be such a sequence.)