Not an answer but a simple heuristic argument: if you set $r=(ab)_p$, the OP's sum is equal to $$\sum_{1\le r\le p-1}\dfrac{1}{r}\sum_{1\le a\le p-1}\dfrac{1}{a}(ra^{-1})_p$$ This proves immediately that the sum is less than $(p-1)H_{p-1}^2$, asymptotically $p\log(p)^2$, and if we assume (heuristic part) that $(ra^{-1})_p$ has average $(p-1)/2$ we indeed obtain a guess of $p\log(p)^2/2$. Maybe this last part can be made rigorous.

Not an answer but a simple heuristic argument: if you set $r=(ab)_p$, the OP's sum is equal to $$\sum_{1\le r\le p-1}\dfrac{1}{r}\sum_{1\le a\le p-1}\dfrac{1}{a}(ra^{-1})_p$$ This proves immediately that the sum is less than $(p-1)H_{p-1}^2$, asymptotically $p\log(p)^2$, and if we assume (heuristic part) that $(ra^{-1})_p$ has average $(p-1)/2$ we indeed obtain a guess of $p\log(p)^2/2$. Maybe this last part can be made rigorous.

EDIT: if you consider the much simpler SINGLE sum $S(p)=\sum_{1\le a\le p-1}\dfrac{(a^{-1})_p}{a}$, the same heuristic would give an asymptotic of $p\log(p)/2$. However, numerically $S(p)/(p\log(p))$ does NOT seem to tend to a limit, but oscillates between something like $0.38$ and $0.52$. This should be much easier to analyze, and perhaps indicate that there is also some oscillation in the OP's original question, with no limit.