In the prime number theorem, the main difficulty is that it is hard to eliminate the case where the limit does not exist and instead oscillates between some values, i.e., a lot of effort is needed in justifying your first statement $\displaystyle \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =c$ You can get bounds for $A = \limsup_{x \to \infty} \dfrac{\psi(x)}{x}$ and $a = \liminf_{x \to \infty} \dfrac{\psi(x)}{x}$. In fact, Selberg proved that (before his elementary proof with Erdos) that $A+a = 2$. The crux in the case of PNT is showing that $A = a$.
Once you prove that this limit exists, then it is relatively easy to get its value to be $1$.
EDIT
A better way to write out what you have written would be as follows:
First prove that
\begin{align*} \sum_{d \leq N} \dfrac{\Lambda(d)}d &= \log N + \mathcal{O}(1) \end{align*} The proof for this goes as follows. We have that $\log(N!) = N \log N + \mathcal{O}(N)$. Also, \begin{align} \log(N!) & = \sum_{d \leq N} \Lambda(d) \left \lfloor \dfrac{N}d \right \rfloor = \sum_{d \leq N} \Lambda(d) \left( \dfrac{N}d + \mathcal{O}(1)\right)\\ & = N \sum_{d \leq N} \dfrac{\Lambda(d)}d + \mathcal{O} \left(\sum_{d \leq N} \Lambda(d)\right) = N \sum_{d \leq N} \dfrac{\Lambda(d)}d + \mathcal{O} \left(N\right) \end{align} Hence, $\sum_{d \leq N} \dfrac{\Lambda(d)}d = \log N + \mathcal{O}(1)$ Now use Abel summation technique or by writing $\sum_{d \leq N} \dfrac{\Lambda(d)}d$ as $\displaystyle \int_{2^-}^x \dfrac{\psi(t)}{t}dt$ and performing integration by parts to conclude that to conclude that if $\lim_{x \to \infty} \dfrac{\psi(x)}x = c$ exists, then $c=1$.