The result is fairly elementary. Lets prove it now:
Recall some common definitions: Let $\theta(x)=\sum_{p\leq x} \log p$, let $\Lambda(n)$ be the Von Mangoldt lambda function and let $\psi(x)=\sum_{n\leq x} \Lambda(n)$
Then as $\log n =\sum_{d|n} \Lambda(n)$ we see that $$\sum_{n\leq x} \log n=\sum_{n\leq x}\sum_{d|n} \Lambda(d)=\sum_{d\leq x} \Lambda(d)\lfloor x/d\rfloor =x\sum_{d\leq x} \frac{\Lambda(d)}{d}+O\left(\psi(x) \right)$$
Lemma 1: $\theta(x)\leq 3x$
The proof of this is postponed until the end.
Now, since $\theta(x)=\psi (x)+O(\sqrt{x})$, we have $\psi(x)=O(x)$. Combining this with the fact that $\sum_{n\leq x} \log n= x\log x + O(x)$ we see that $$\sum_{d\leq x} \frac{\Lambda(d)}{d}=\log x+O(1).$$ Since the sum over all the prime powers of $\frac{\Lambda(d)}{d}$ is bounded by a constant, we conclude $$\sum_{p\leq x}\frac{\log p}{p}=\log x+O(1).$$
Hope that helps,
Proof of Lemma 1:
Consider the binomial coefficient $$\binom{2N}{N}.$$ Notice that every prime in the interval $(N,2N]$ appears in the numerator. Then $$\prod_{N<p\leq2N}p\leq\binom{2N}{N}$$ Since $$\binom{2N}{N}\leq(1+1)^{2N}=4^{N},$$ we see that $$\prod_{N<p\leq2N}p\leq4^{N},$$ and hence $\theta(2N)-\theta(N)\leq N\log4.$ Consider $N$ of the form $N=2^{r}.$ Then we get the list of inequalities: $$\theta\left(2\right)-\theta\left(1\right)\leq\log4$$
$$\theta\left(4\right)-\theta\left(2\right)\leq2\log4$$
$$\theta\left(8\right)-\theta\left(4\right)\leq4\log4$$
$$\theta(2^{r})-\theta\left(2^{r-1}\right)\leq2^{r-1}\log4$$
$$ \theta\left(2^{r+1}\right)-\theta\left(2^{r}\right)\leq2^{r}\log4.$$
Summing all of these yields $$\theta\left(2^{r+1}\right)-\theta\left(1\right)\leq\left(2^{r}+\cdots+1\right)\log4\leq2^{r+1}\log4$$ for each $r$. For every $x$ in the interval $(2^{r},2^{r+1}]$ we have $\theta(x)\leq\theta(2^{r+1})$ so that $$\theta(x)\leq2^{r+1}\log4\leq x\cdot2\log4<3x$$ since $x>2^{r}$. Thus the result is proven.
We have that $$\pi_k(x)\sim\sum_{p_k\leq x}1 \sim \frac{x (\log \log x)^{k-1}}{(k-1)!\log x}\ \ \ \ \ \ \ \ \ \ (A)$$ and $$\vartheta_k(x)\sim\sum_{p_k\leq x}\log (p_k) \sim \frac{x (\log \log x)^{k-1}}{(k-1)!},\ \ \ \ \ \ \ \ \ \ (B)$$ so I can reinterpret your question as how does $(B)$ follow from $(A)$, as that implies your equation $2$. Equivalently, you are asking, if I know $(A)$, can I prove that $$\theta_k(x)\sim \pi_k(x)\log x.$$ The short answer is that this follows directly from partial summation. First, lets write the sum as a Riemann Stieltjes integral $$\sum_{p_k\leq x} \log p_k=\int_1^x \log t d\left(\pi_k(t)\right),$$ and then using integration by parts, this equals $$\pi_k(x)\log x-\int_{1}^x\frac{\pi_k(t)}{t}dt.$$ Now, it is possible to use the asymptotic $(A)$ to bound the second integral from above, and we have reduced your question to an elementary real analysis problem.
You may also want to take a look at the paper: A simple proof of a theorem of Landau by E.M. Wright. This paper gives an elementary proof of $(A)$ using the hyperbola method.
Additionally, here is a note related problem which I wrote at some point. It may interest you.
Best Answer
(a) $\to$ (b). I show one half of part (b), namely $\liminf \rho(x) \leqslant 1$. The other half is analogous.
Assume that there exists $\delta >0$ and $T \geqslant 2$ such that $\rho(t) \geqslant 1+\delta$ for all $t \geqslant T$. That is, $$ \pi(t) \geqslant (1+ \delta) \frac{t}{\ln t}, \tag{$\ast$} $$ for all $t \geqslant T$. Therefore, $$ \begin{align*} \int_2^x \frac{\pi(t)}{t^2} \, dt & \geqslant \int_T^x \frac{\pi(t)}{t^2} \, dt \\ &\stackrel{(\ast)}{\geqslant} \int_T^x (1 + \delta) \frac{t}{\ln t} \cdot \frac{1}{t^2} \, dt \\ &=(1 + \delta) \int_T^x \frac{1}{t \, \ln t} \, dt \\ &= (1 + \delta) \cdot \left. \ln \ln t \right\vert_{t=T}^{t=x} \\ &= (1 + \delta) \cdot (\ln \ln x - \ln \ln T) \\ &\sim (1 + \delta) \cdot \ln \ln x, \end{align*} $$ for large $x$. This clearly contradicts part (a). It thus follows that $\liminf \rho(x) \leqslant 1$.