Partial cover approximation
Having discussed with other users, I believe the following answers the question. 

To begin with let us recap the greedy algorithm for the set cover problem, in which we wish to cover all of $E$ as cheaply as possible. At each step, the algorithm adds the set $S_i$ for which $\frac{w_i}{|\hat{S}_i|}$ is lowest, where $\hat{S}_i$ is the subset of $S_i$ which has yet to be covered. It can be shown that this minimum is upper bounded by $\frac{OPT}{Z}$ where $OPT$ is the optimal weight for the (full) set cover problem and $Z$ is the number of elements not yet covered. From this, one can show a a total weight bounded by $H_{|E|} \approx \log |E|$. 

However, if one were to stop the algorithm as soon as $p|E|$ elements were covered, then we would get the bound  $\approx \log \frac{1}{1-p}$, as per @SashoNikolov's comment.

We can get a $H_{p|E|}$ approximation for the partial set problem by modifying the greedy algorithm as follows: At each iteration, choose the $S_i$ which minimizes $\frac{w_i}{\min(|\hat{S}_i|, Z)}$ where $Z$ is the number of elements we still need to get to complete $p|E|$ elements. Now we can guaranteed that there is an $S_i$ for which the above is bounded by $OPT/Z$ (where now $OPT$ is the optimal for the partial set cover). To see this, suppose an optimal solution is the collection of sets $S_1=\{xx|...\}$ $S_2=\{xxx|...\}$, $S_3=\{xxx|......\}$. In this notation, the $x$ represents an already covered element, and after the $|$ are elements which are not covered. Let $a_1, a_2, a_3$ denote the number of such uncovered elements and suppose $a_1, a_2, a_3\leq Z$. Then 
$$min \{\frac{w_1}{a_1}, \frac{w_2}{a_2}, \frac{w_3}{a_3}\}\leq \frac{w_1+w_2+w_3}{a_1+a_2+a_3}  = \frac{OPT}{a_1+a_2+a_3} \leq \frac{OPT}{Z}.$$

Of course, if any of the $a_i$'s is larger than $Z$ then it is replaced by $Z$ in the above inequalities and they hold.

The starting $Z$ is $Z_0=p|E|$, and a similar analysis as for the full set cover leads to a total weight bounded by $OPT \cdot H_{Z_0}$.