The sum on the left is $ \sum_{i,j=1}^{\infty} \left[ \frac x{p_i p_j} \right] = \sum_{p_i p_j \le x} \left[ \frac{x}{p_i p_j} \right] $ because if $p_i p_j > x$ then the integer part is just $0$, hence you don't need to sum over infinitely many terms. It remains to see that there is $O(x)$ terms in that sum, since $ z = [z] + \{z\} \qquad \text{ with } \qquad \{z\} \le 1. $ so that you can look at the sum on the left as the sum on your right + the remainders, which you will only need to "count" how many of them appear, since all the terms you will sum will be less than $1$. Hence showing that there is at most $O(x)$ terms in that sum shows that you can bound the remainder's sum by $O(x) \times 1 = O(x)$.
The number of terms in that sum is $\begin{align*} \# \left\{ (p_i,p_j) \, | \, p_i p_j \le x, p_i < p_j \right\} &\le \, \# \left \{ n \, | \, n \text{ is composite }, n \le x \right \} \\ &= x - \pi(x) \\ &\sim x - \frac {x}{\log x} \\ &= O(x). \end{align*}$ (A rough bound though would just be $x$ and it would still be $O(x)$, I just thought I'd do a little better.)
(ADDED : Aleks was wondering why my inequality held, so I'll add the proof here. Call the set with the prime couples $A$ and the set with the composites $B$. The application which takes a couple $(p_i, p_j)$ with $p_i < p_j$ to their product $p_i p_j = n$ is injective, for if $p_i^1 p_j^1 = n_1 = n_2 = p_i^2 p_j^2$, since $p_i^k < p_j^k$ we must have $p_i^1 = p_i^2$ and $p_j^1 = p_j^2$. Since this application is injective, $\# A \le \# B$. )
This gives you $\begin{align*} \sum_{i,j=1}^{\infty} \left[ \frac{x}{p_i p_j} \right] &= \sum_{p_i p_j \le x} \left[ \frac x{p_i p_j} \right]\\ &= \sum_{p_i p_j \le x} \left( \frac x{p_i p_j} - \left\{ \frac{x}{p_ip_j} \right\} \right)\\ &= x \left( \sum_{p_i p_j \le x} \frac 1{p_i p_j} \right) - \sum_{p_i p_j \le x} \left\{ \frac{x}{p_ip_j} \right\} \end{align*}$ and since the last sum contains $O(x)$ terms less than $1$, it is $O(x)$.
Hope that helps,