Well we know that
$\zeta(\sigma)= \frac{1}{\sigma-1}$ when $ \sigma \to1^+$ since zeta has a simple pole in $\sigma=1$ with residual one. Now your function $\sum_p\frac{\log p}{p^s}$ can be expressed in terms of the zeta function.
see http://people.math.jussieu.fr/~demarche/enseignements/2011-2012/M1-TDN/MM020-TD6-corrige.pdf (it's the exercise number 7, question g, in french :)). The tricky part is why could you integrate two functions which are equivalent in the sense that $\sum_p \frac{\log p}{p^s} ∼ \frac{1}{s-1}$ ?
Here it is possible since $1/s-1$ is not integrable in a neighborhood of 1. Why?
here is an sketch of how you could procede. Let f, g be continuous on R such that f~g and $\int_0^\infty$fdx diverges. I will write this integral I in what follows and I$x$ when the upper bound is $x$ instead of + $\infty$. we would like to show that I (f) ~ I (g)
In what follows we have f ~ g at infinity. (I guess you could repeat that proof for an equivalence that holds in another neighborhood)
f ~ g then f (x) = g (x) (1 + h (x)) such that $gh \to0$ when x tends to + infinity (and h tends to $0$ as well).
So you've got I$x$ (f) / I$x$ (g) = 1 + I$x$ (gh) / I$x$ (g) then show that | I$x$ (gh) / I$x$ (g) | tends to $0$ in + infinity
since |I$x$ (gh) / I$x$ (g) | = I$a$ (gh) / I$x$ (g) + I$a-x$ (gh) / Ix (g) then | I$x$ (gh) / I$x$ (g) | $ \leq$ I$a$(gh) / I$x$ (g) + sup (h) on [$a$, x]
and since h tends to $0$, you can choose $a$ large enough such that $sup (h) \leq \epsilon / 2$
Once $a$ is chosen, since I$a$ (gh) is a number and I$x$ (g) diverges then there exists $x$ large enough such that I$a$ (gh) / I$x$ (g) $\leq \epsilon / 2$. Finally, for $x$ large enough everything is inferior to a certain epsilon, and you can conclude.