A few years ago, I assigned something a little weaker, $$\sum_{n\le x}{\Lambda(n)\over n}=\log x+r(x){\rm\ with\ }|r(x)|\le2$$ and it turned out to be the hardest problem on the assignment. Here's the proof I eventually wrote up.
Let $s=[x]$. Note $$\sum_{n\le x}{\Lambda(n)\over n}=\sum_{n\le s}{\Lambda(n)\over n}$$ and $0\le\log x-\log s\lt1$. Now
$$\log s!=\sum_{m\le s}\log m=\sum_{p^r\le s}\log p\sum_{m\le s,p^r\mid m}1=\sum_{p^r\le s}\log p[s/p^r]=\sum_{n\le s}\Lambda(n)[s/n]$$
$$=\sum_{n\le s}\Lambda(n)\left({s\over n}-\left\lbrace{s\over n}\right\rbrace\right)=s\sum_{n\le s}{\Lambda(n)\over n}-\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace$$
Now $$\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace\le\sum_{n\le s}\Lambda(n)\le Cs$$ for some constant $C$ by the Prime Number Theorem (which is actually overkill, as one can prove without PNT that that last sum is bounded by $2s$ for all $s$). Also, comparing $\log s!=\sum\log n$ to $\int_1^s\log t\,dt$ we get $$\log s!=s\log s-s+b(s){\rm\ with\ }|b(s)|\lt\log s+1$$ (cf. Stirling's formula). Putting it together, $$\sum_{n\le x}{\Lambda(n)\over n}=(1/s)\log s!+(1/s)\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace=\log s-1+C+b(s)/s=\log x+r(x)$$ with $|r(x)|\le2$ since $C\le2$.
EDIT: The stronger result is proved in the textbook by Bateman and Diamond, Analytic Number Theory, pages 100-102. First they prove that, with the usual definitions, $\psi(x)\sim x$ implies $M(x)=o(x)$, where $M(x)$ is the summatory function for the Möbius $\mu$-function. Then from the result on $M(x)$ they deduce the result you want. It's a bit long to write out as I'd have to explain the notation introduced earlier in the chapter.
MORE EDIT: It's also Proposition 3.4.4 of Jameson's textbook, The Prime Number Theorem, and it's proved on page 91 of Tenenbaum and Mendes France, The Prime Numbers and Their Distribution, in the Student Mathematical Library series of the American Math Society. Again, the proofs require too much previously developed material for me to attempt to write them out here.
Here is a general strategy. You are interested in
$$\sum_{p \leq x} \frac{1}{p^{1 + \frac{2}{\log x}}} - \frac{2}{\log x} \sum_{p \leq x} \frac{\log p}{p^{1 + \frac{2}{\log x}}} + \frac{1}{(\log x)^2} \sum_{p \leq x} \frac{(\log p)^2}{p^{1 + \frac{2}{\log x}}}.$$
We use partial summation, noting that
$$\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right), \qquad \sum_{p \leq x} \frac{\log p}{p} = \log x + O(1), \qquad \sum_{p \leq x} \frac{(\log p)^2}{p} = \frac{(\log x)^2}{2} + O(1).$$
Here $b$ is some explicit constant.
The first terms via partial summation are
$$e^{-2} \log \log x + b e^{-2}, \qquad -2e^{-2}, \qquad \frac{e^{-2}}{2}.$$
The error is $O(1/\log x)$ for the first two and $O(1/(\log x)^2)$ for the third.
For the second terms, we note that
$$\frac{d}{dt} \frac{1}{t^{\frac{2}{\log x}}} = -\frac{2}{\log x} \frac{1}{t} \exp\left(-\frac{2 \log t}{\log x}\right),$$
and so we must evaluate
$$\frac{2}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{2b}{\log x} \int_{2}^{x} \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad -\frac{4}{(\log x)^2} \int_{2}^{x} \log t \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}, \qquad \frac{1}{(\log x)^3} \int_{2}^{x} (\log t)^2 \exp\left(-\frac{2\log t}{\log x}\right) \, \frac{dt}{t}.$$
(There are also error terms; the method below shows that they are $O(\log \log x/\log x)$.) We make the change of variabes $t \mapsto e^t$, so that these become
$$\frac{2}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{2b}{\log x} \int_{\log 2}^{\log x} \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad -\frac{4}{(\log x)^2} \int_{\log 2}^{\log x} t \exp\left(-\frac{2 t}{\log x}\right) \, dt, \qquad \frac{1}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
The latter three can be explicitly evaluated (directly for the second, via integration by parts for the last two); they give
$$-b e^{-2} - b \exp\left(-\frac{2 \log 2}{\log x}\right), \qquad 3e^{-2} - \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 \log 2}{\log x} + 1\right), \qquad -\frac{5}{4} e^{-2} + \frac{1}{4} \exp\left(-\frac{2 \log 2}{\log x}\right) \left(\frac{2 (\log 2)^2}{(\log x)^2} + \frac{2 \log 2}{\log x} + 1\right).$$
Note that $\exp(-2\log 2/\log x) = 1 + O(1/\log x)$. For the former, we integrate by parts once, getting
$$-e^{-2} \log \log x + \log \log 2 \exp\left(-\frac{2 \log 2}{\log x}\right) + \int_{\log 2}^{\log x} \frac{1}{t} \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
For this second term, we make the change of variables $t \mapsto \frac{\log x}{2} t$, so that this becomes
$$\int_{\frac{2 \log 2}{\log x}}^{2} \frac{e^{-t}}{t} \, dt = E_1\left(\frac{2 \log 2}{\log x}\right) - E_1(2),$$
where $E_1(z)$ is the exponential integral. Since $E_1(z) = -\log z - \gamma_0 + O(z)$ about $z = 0$, we get additional terms
$$\log \log x - \log 2 - \log \log 2 - \gamma_0 - E_1(2) + O\left(\frac{1}{\log x}\right).$$
So now we combine everything and find that the desired asymptotic is
$$\log \log x - b + \frac{e^{-2}}{4} + \frac{1}{4} - \log 2 - \gamma_0 - E_1(2) + O\left(\frac{\log \log x}{\log x}\right).$$
Alternatively, you can use your approach to get the expression
$$\frac{4}{\log x} \int_{2}^{x} \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} - \frac{6}{(\log x)^2} \int_{2}^{x} \log t \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t} + \frac{2}{(\log x)^3} \int_{2}^{x} (\log t)^2 \log \log t \exp\left(-\frac{2 \log t}{\log x}\right) \, \frac{dt}{t}.$$
(Note that you seem to have calculated the derivative incorrectly.) We again make the change of variables $t \mapsto e^t$, yielding
$$\frac{4}{\log x} \int_{\log 2}^{\log x} \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt - \frac{6}{(\log x)^2} \int_{\log 2}^{\log x} t \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt + \frac{2}{(\log x)^3} \int_{\log 2}^{\log x} t^2 \log t \exp\left(-\frac{2 t}{\log x}\right) \, dt.$$
Now integrate by parts repeatedly, integrating the exponential and differentiating everything else.
Best Answer
If $$\sum_{n \le x} f(n)=O(x^a \log^b x)$$ then $$\sum_{n \le x} \frac{f(n)}{\log n} =\frac1{\log x} \sum_{n \le x} f(n)+\sum_{n \le x-1}( \sum_{m \le n} f(m)) (\frac1{\log n}-\frac1{\log (n+1)} )$$ $ \frac1{\log n}-\frac1{\log (n+1)} = \frac{\log(n+1)-\log n}{\log n \log (n+1)} = O(\frac1{n \log^2 n})$ $$ = O( x^a \log^{b-1} x+\sum_{n \le x} n^{a-1} \log^{b-2} n) =O( x^a \log^{b-1} x+\log^{b-2} x\sum_{n \le x} n^{a-1} )=O(x^a \log^{b-1} x)$$
Doing it with $f(n) = \Lambda(n) 1_{n \equiv a \bmod q}- \frac{\Lambda(n)}{\phi(q)}$ and using $\sum_{p^k \le x, k \ge 2} \frac1k (1_{p^k\equiv a \bmod q}-\frac1{\phi(q)}) = O(x^{1/2})$ gives the result