Linear regression suggests that there are $a,b\in\mathbb R^+$ such that
$$\displaystyle a+bx\sim \frac{x\lfloor e^x\rfloor}{p_{n}-x\lfloor e^x\rfloor}\tag1,\; n=\lfloor e^x\rfloor$$
where $p_n$ is the $n$-th prime and $\sim$ denotes asymptotically equivalence. Can this can be proved?
0 0.00000
1 9.64894
2 6.65115
3 5.60872
4 6.69764
5 6.45644
6 6.98725
7 6.81400
8 7.09048
9 7.28458
10 7.46297
11 7.62874
12 7.86839
13 8.08110
14 8.32635
15 8.55811
16 8.80205
17 9.04825
18 9.29361
19 9.54235
20 9.79004
21 10.03895
22 10.28804
23 10.53697
24 10.78539
25 11.03331
26 11.28060
27 11.52714
28 11.77287
29 12.01775
30 12.26173
31 12.50480
It follows from the prime number theorem that $$\displaystyle \lim_{n\to\infty}\frac{p_n}{n\ln n}=1\tag2$$ Thus, for any function $f(n)$ asymptotically equivalent with $0$, it holds that $$\displaystyle \lim_{n\to\infty}\frac{p_n}{(1+f(n))\cdot n\ln n}=1\tag3$$ If there exist $a$ and $b$ as above, select $$\displaystyle f(n)=\frac{1}{a+b\cdot\ln n}\tag4$$ would give a rather accurate smooth approximation $$\displaystyle p_n\approx \Big(1+\frac{1}{a+b\cdot\ln n}\Big)\cdot n\ln n\tag5$$
Addendum:
Extending the computations and making linear regression, strongly suggest that $a$ and $b$ exists and that $a\approx 4.84615$, $b\approx 0.247308$ and obviously (1) implies (5).
If $(5)$ is meant to be accurate enough to imply $$\lim_{n\to\infty}\log n\left(\frac{p_n}{n\log n}-1\right)=\frac1b,$$then your conjecture is false. This follows from the well known lower bound$$\frac{p_n}{n}>\log n+\log\log n-1.$$