The leading term of the asymptotics is $\frac{\pi^2}{12}\cdot \frac{x^2}{\log x}$. The difference to your empirical estimate $c \approx 0.86$ comes from the lower order terms, mainly the $O\Bigl(\frac{x^2}{\log^2 x}\Bigr)$ term. I shall only prove the leading term, but with more work further terms can be identified.
Writing $p^k \mathrel{\Vert} n$ for $p^k \mid n$, $p^{k+1}\nmid n$, changing the order of summation yields
\begin{align}
\sum_{n \leqslant x} \sum_{p^k \mathrel{\Vert} n} p^k
&= \sum_{p^k \leqslant x} p^k\cdot \varphi(x/p^k,p) \\
&= \sum_{p^k \leqslant x} p^k\biggl(\biggl\lfloor \frac{x}{p^k}\biggr\rfloor
- \biggl\lfloor \frac{x}{p^{k+1}}\biggr\rfloor\biggr)\,,
\end{align}
where $\varphi(y,m)$ is the number of positive integers not exceeding $y$ that are coprime to $m$.
For $2 \leqslant k \leqslant \frac{\log x}{\log 2}$, the contribution to the total is easily seen to be $O(x^{3/2})$ – each term in the sum is $\leqslant x$, and there are only $\sum_{k \geqslant 2} \pi(x^{1/k}) = o(\sqrt{x})$ terms –, so we need only consider $k = 1$.
Now
\begin{equation}
\sum_{p \leqslant x} p\biggl\lfloor \frac{x}{p^2}\biggr\rfloor
\leqslant x\sum_{p \leqslant x} \frac{1}{p} = O(x\log\log x)\,,
\end{equation}
hence the only relevant bit is
\begin{align}
\sum_{p \leqslant x} p\biggl\lfloor \frac{x}{p}\biggr\rfloor
&= \sum_{p\cdot m \leqslant x} p\cdot 1 \\
&= \sum_{m \leqslant \sqrt{x}} S\biggl(\frac{x}{m}\biggr)
+ \sum_{p \leqslant \sqrt{x}} p\biggl\lfloor \frac{x}{p}\biggr\rfloor
- \lfloor \sqrt{x}\rfloor S(\sqrt{x}) \\
&= \sum_{m \leqslant \sqrt{x}} S\biggl(\frac{x}{m}\biggr) + x\pi(\sqrt{x}) + O\bigl(S(\sqrt{x})\bigr)
- \lfloor \sqrt{x}\rfloor S(\sqrt{x}) \\
&= \sum_{m \leqslant \sqrt{x}} S\biggl(\frac{x}{m}\biggr) +O(x^{3/2})
\end{align}
where $S(y) = \sum_{p \leqslant y} p$, using the trivial estimates
$\pi(y) \leqslant y$ and $S(y) \leqslant y\cdot\pi(y) = O(y^2)$.
For the remaining sum we need
\begin{equation}
\DeclareMathOperator{\Li}{Li}
S(y) = \Li(y^2) + O\biggl(\frac{y^2}{\log^A y}\biggr) \tag{$\ast$}
\end{equation}
for every $A > 0$ (if we want an arbitrary number of terms; for $k$ terms, any $A > k$ suffices). This follows from $\pi(x) = \Li(x) + O(x/\log^A x)$ by partial summation:
\begin{align}
\sum_{p \leqslant x} p &= x\pi(x) - \int_2^x \pi(u)\,du \\
&= x\Li(x) - \int_2^x \Li(u)\,du + O\Biggl(\frac{x^2}{\log^A x} + \int_2^x \frac{u}{\log^A u}\,du\Biggr) \\
&= \int_2^x \frac{u}{\log u}\,du + O\biggl(\frac{x^2}{\log^A x}\biggr) \\
&= \int_2^x \frac{2u}{\log (u^2)}\,du + O\biggl(\frac{x^2}{\log^A x}\biggr) \\
&= \Li(x^2) - \Li(4) + O\biggl(\frac{x^2}{\log^A x}\biggr)\,.
\end{align}
Since $\frac{1}{2}\log x \leqslant \log \frac{x}{m} \leqslant \log x$ for $1 \leqslant m \leqslant \sqrt{x}$,
\begin{equation}
\sum_{m \leqslant \sqrt{x}} \frac{x^2}{m^2\log^A \frac{x}{m}}
\leqslant \frac{2^Ax^2}{\log^A x}\sum_{m \leqslant \sqrt{x}} \frac{1}{m^2} = O\biggl(\frac{x^2}{\log^A x}\biggr)\,.
\end{equation}
Then using the asymptotic expansion
\begin{equation}
\Li(y) = \sum_{k = 1}^{\ell} \frac{y\cdot (k-1)!}{\log^{k} y} + O\biggl(\frac{y}{\log^{\ell+1} y}\biggr)
\end{equation}
with $y = \frac{x^2}{m^2}$ and arbitrary $\ell$ yields
\begin{align}
\sum_{m \leqslant \sqrt{x}} S\biggl(\frac{x}{m}\biggr)
&= \sum_{m \leqslant x} \Biggl(\sum_{k = 1}^{\ell}
\frac{x^2(k-1)!}{2^k m^2 \log^k \frac{x}{m}} + O\biggl(\frac{x^2}{m^2\log^{\ell + 1} \frac{x}{m}}\biggr)\Biggr)
+ O\biggl(\frac{x^2}{\log^A x}\biggr) \\
&= \sum_{k = 1}^{\ell} \frac{x^2(k-1)!}{2^k}
\sum_{m\leqslant \sqrt{x}} \frac{1}{m^2\log^k \frac{x}{m}}
+ O\biggl(\frac{x^2}{\log^{\min \{A,\ell+1\}} x}\biggr)\,.
\end{align}
Now
\begin{equation}
\frac{1}{\log^k \frac{x}{m}} = \frac{1}{\log^k x\bigl(1 - \frac{\log m}{\log x}\bigr)^k}
= \sum_{\nu = 0}^{r} (-1)^{\nu}\binom{-k}{\nu} \frac{\log^{\nu} m}{\log^{k + \nu} x} + O\biggl(\frac{\log^{r+1}m}{\log^{k+r+1} x}\biggr)\,,
\end{equation}
$(-1)^{\nu}\binom{-k}{\nu} = \binom{k-1+\nu}{\nu}$ and
\begin{equation}
\sum_{m \leqslant z} \frac{\log^{\nu} m}{m^2} = (-1)^{\nu}\zeta^{(\nu)}(2) + O\biggl(\frac{\log^{\nu} z}{z}\biggr)\,,
\end{equation}
whence by choosing $A = \ell + 1$ we obtain
\begin{align}
\sum_{m \leqslant \sqrt{x}} S\biggl(\frac{x}{m}\biggr)
&= \sum_{k = 1}^{\ell} \frac{x^2(k-1)!}{2^k}
\Biggl(\sum_{\nu = 0}^{\ell-k}
\frac{(k-1+\nu)!}{\nu!(k-1)!\log^{k + \nu} x}
\sum_{m \leqslant \sqrt{x}} \frac{\log^{\nu} m}{m^2}
+ O\bigl(\log^{-\ell-1} x\bigr)\Biggr) \\
&\qquad\qquad + O\biggl(\frac{x^2}{\log^{\ell+1} x}\biggr) \\
&= \sum_{k = 1}^{\ell}\sum_{\nu =0}^{\ell-k} \frac{x^2(k-1+\nu)!}{2^k\nu!\log^{k+\nu} x}\biggl((-1)^{\nu}\zeta^{(\nu)}(2) + O\biggl(\frac{\log^{\nu} \sqrt{x}}{\sqrt{x}}\biggr)\biggr)
+ O\biggl(\frac{x^2}{\log^{\ell + 1} x}\biggr) \\
&= \sum_{\mu = 1}^{\ell} \frac{x^2}{\log^{\mu} x}
\sum_{\nu = 0}^{\mu-1} \frac{(-1)^{\nu}\zeta^{(\nu)}(2)(\mu-1)!}{2^{\mu-\nu}\nu!} + O\biggl(\frac{x^2}{\log^{\ell + 1} x}\biggr)\,.
\end{align}
Expanding the first few inner sums leads to the asmptotics
\begin{align}
\frac{\zeta(2)}{2}\cdot \frac{x^2}{\log x}
&+ \biggl(\frac{\zeta(2)}{4} - \frac{\zeta'(2)}{2}\biggr)
\frac{x^2}{\log^2 x}
+ \biggl(\frac{\zeta(2)}{4} - \frac{\zeta'(2)}{2}
+ \frac{\zeta''(2)}{2}\biggr)\frac{x^2}{\log^3 x} \\
&+ \biggl(\frac{3\zeta(2)}{8} - \frac{3\zeta'(2)}{4}
+ \frac{3\zeta''(2)}{4} - \frac{\zeta'''(2)}{2}\biggr)
\frac{x^2}{\log^4 x} \\
&+ \biggl(\frac{3\zeta(2)}{4} - \frac{3\zeta'(2)}{2}
+ \frac{3\zeta''(2)}{2} - \zeta'''(2)
+ \frac{\zeta^{(4)}}{2}\biggr)\frac{x^2}{\log^5 x}
+ O\biggl(\frac{x^2}{\log^6 x}\biggr)\,.
\end{align}
