The summation by parts yields
$$\sum_{n \leqslant x} \frac{\Lambda(n)}{n} = \frac{\psi(x)}{x} + \int_1^x \frac{\psi(t)}{t^2}\,dt = \log x + \frac{\psi(x)}{x} + \int_1^x \frac{\psi(t)-t}{t^2}\,dt\,.$$
Thus it is easy to see that under the hypothesis $\psi(x) \sim x$ the limit
$$\lim_{x\to \infty}\; \Biggl(\log x - \sum_{n \leqslant x} \frac{\Lambda(n)}{n}\Biggr) \tag{1}$$
exists if and only if
$$\lim_{x \to \infty} \int_1^{x} \frac{\psi(t) - t}{t^2}\,dt \tag{2}$$
exists. It is easy to see that if $(1)$ exists, then the limit is $\gamma$, so it would suffice to prove the existence of $(1)$ without caring about its value.
Actually, one step in Newman's proof of the prime number theorem is to show that the existence of $(2)$ implies $\psi(x) \sim x$ (see e.g. here; although Zagier's exposition of Newman's proof uses $\vartheta$ rather than $\psi$, the argument is practically identical for both choices), so the existence of $(2)$ implies that of $(1)$ without further assumptions. And since the existence of $(1)$ implies $\psi(x) \sim x$ via summation by parts, we see that $(1)$ and $(2)$ are elementarily equivalent. However, on the face of it, $(2)$ seems to be stronger than $\psi(x) \sim x$, and there's no obvious direct way to obtain $(2)$ from $\psi(x) \sim x$. Plugging $\psi(t) = t + o(t)$ into the integral cannot yield more than
$$\int_1^x \frac{\psi(t) - t}{t^2}\,dt \in o(\log x)\,.$$
We need a different angle of attack. Until yesterday evening, I was not aware of a direct elementary way to show that $\psi(x) \sim x$ implies the existence of $(1)$, and my answer would have been to first show (using Landau's proof) that $\psi(x) \sim x$ implies the convergence of
$$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}$$
and then use that to reach the goal. This route is fun, but it is also rather long. But yesterday evening, I happened to see this Math Overflow answer by Vesselin Dimitrov sketching a short elementary and direct proof of
$$\psi(x) \sim x \implies \sum_{n \leqslant x} \frac{\Lambda(n)}{n} = \log x - \gamma + o(1)\,.$$
This is very much a fun proof too, so let's fill in the details.
First, we define
$$\delta(x) := \sup\; \biggl\{ \biggl\lvert \frac{\psi(y)}{y} - 1\biggr\rvert : y \geqslant x\biggr\}$$
for $x\geqslant 1$. Then $\delta$ is nonicreasing, and $\delta(x) \to 0$ for $x \to \infty$ is just our assumption $\psi(x) \sim x$. I note that $\lvert\delta(x)\rvert \leqslant 1$ for all $x \geqslant 1$. Then we will need a function $\omega \colon [1,\infty) \to [1,\infty)$ such that $\lim\limits_{x\to\infty} \omega(x) = \infty$ and
$$\lim_{x\to \infty} \delta(x)\log \omega(x) = 0\,.$$
We can take $\omega(x) = 1/\delta(x)$ for example. Next, following Chebyshev, let
$$T(x) := \sum_{n \leqslant x} \log n = \log \bigl(\lfloor x\rfloor!\bigr)\,.$$
From Euler's sum formula, summation by parts, or Stirling's formula one easily obtains
$$\lvert T(x) - (x\log x - x)\rvert \leqslant 1 + \log x$$
for $x \geqslant 1$, and writing Legendre's formula in terms of the von Mangoldt function, we have
$$T(x) = \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{x}{n}\biggr\rfloor\,.$$
Now put $z = x\omega(x)$ and by splitting the sum at $x$ find
\begin{align}
T(z) &= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor
+ \sum_{x <n \leqslant z} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor \\
&= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor
+\sum_{x < n \leqslant z} \Lambda(n)\sum_{k \leqslant z/n} 1 \\
&= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor
+ \sum_{k < \omega(x)} \sum_{x < n \leqslant z/k} \Lambda(n) \\
&= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor
+ \sum_{k \leqslant \omega(x)} \Biggl(\psi\biggl(\frac{z}{k}\biggr) - \psi(x)\Biggr) \\
&= z\sum_{n \leqslant x} \frac{\Lambda(n)}{n}
+ \sum_{k \leqslant \omega(x)} \Biggl(\psi\biggl(\frac{z}{k}\biggr) - \psi(x)\Biggr)
- \sum_{n \leqslant x} \Lambda(n)\biggl\lbrace \frac{z}{n}\biggr\rbrace\,.
\end{align}
Dividing by $z$ and rearranging then yields
\begin{align}
\Biggl\lvert \sum_{n \leqslant x} \frac{\Lambda(n)}{n} &{}- \log x + \gamma \Biggr\rvert \\
&= \Biggl\lvert \frac{T(z)}{z} - \log x + \gamma + \frac{\lfloor \omega(x)\rfloor}{\omega(x)}\cdot \frac{\psi(x)}{x} - \sum_{k \leqslant \omega(x)} \frac{1}{z}\psi\biggl(\frac{z}{k}\biggr) + \frac{1}{z}\sum_{n \leqslant x} \Lambda(n)\biggr\lbrace \frac{z}{n}\biggr\rbrace\Biggr\rvert \\
&\leqslant \Bigl\lvert \log z - 1 - \log x + \gamma + \frac{\lfloor \omega(x)\rfloor}{\omega(x)} - \sum_{k \leqslant \omega(x)} \frac{1}{k}\Biggr\rvert \\
&\qquad + \frac{1 + \log z}{z} + \frac{\lfloor\omega(x)\rfloor}{\omega(x)}\delta(x) + \sum_{k \leqslant \omega(x)} \frac{1}{k}\delta\biggl(\frac{z}{k}\biggr) + \frac{\psi(x)}{z} \\
&\leqslant \biggl\lvert 1 - \frac{\lfloor \omega(x)\rfloor}{\omega(x)}\biggr\rvert + \Biggl\lvert \log \omega(x) + \gamma - \sum_{k \leqslant \omega(x)} \frac{1}{k}\Biggr\rvert \\
&\qquad + \frac{1+\log z}{z} + \delta(x)(2+\log \omega(x)) + \frac{2}{\omega(x)} \\
&\leqslant \frac{4}{\omega(x)} + \frac{1 + \log x + \log \omega(x)}{x\omega(x)} + \delta(x)(2 + \log \omega(x))\,.
\end{align}
The right hand side tends to $0$ for $x \to \infty$ by the constraints imposed on $\omega$.
