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Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

EDIT 1. I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.

EDIT 2. I deleted the comment regarding a possible multiple zero of $\zeta(s)$, since the corresponding zero of $-\zeta'(s)/\zeta(s)=\sum\Lambda(n)n^{-s}$ is always simple.

EDIT 3. Here is a very recent and closely related arXiv preprint.

Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

Added. Here is a closely related arXiv preprint.

Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

EDIT 1. I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.

EDIT 2. I deleted the comment regarding a possible multiple zero of $\zeta(s)$, since the corresponding zero of $-\zeta'(s)/\zeta(s)=\sum\Lambda(n)n^{-s}$ is always simple.

Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

EDIT 1. I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.

EDIT 2. I deleted the comment regarding a possible multiple zero of $\zeta(s)$, since the corresponding zero of $-\zeta'(s)/\zeta(s)=\sum\Lambda(n)n^{-s}$ is always simple.

EDIT 3. Here is a very recent and closely related arXiv preprint.

Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$, while in the case of a multiple root for $\zeta(s)$ we can improve the bounds by the appropriate power of $\ln x$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

EDIT. I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.

Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$ \sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x), $$ then we have, using Riemann-Stieltjes integrals, $$ F(x):=\sum_{x < p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t) $$ $$ = \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x < t \leq 2x} R(t) ). $$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$ S(x):=\sum_{p \leq x} \frac{\ln p}{p} $$ we have $$ \hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0. $$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$ F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}). $$ This implies, by our initial calculation, $$ R(x) = \Omega(x^{\sigma-1}/\ln x). $$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.

EDIT 1. I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.

EDIT 2. I deleted the comment regarding a possible multiple zero of $\zeta(s)$, since the corresponding zero of $-\zeta'(s)/\zeta(s)=\sum\Lambda(n)n^{-s}$ is always simple.