This is not properly an answer, after the comments of Tao it is difficult to give an answer. Only an explanation of my comment above. I still think that my series possibly converges if and only the integral converges and to the same value.

We can write the integral $I$ in the form $$I=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{\zeta(1+ixt)\zeta(1-ixt)}\frac{dt}{t^2}.$$ Hence I consider the function $$u(\sigma)=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{\zeta(\sigma+it)\zeta(\sigma-it)}\frac{dt}{(\sigma-1)^2+t^2},\qquad \sigma>1.$$ I expect to have $\lim_{\sigma\to1^+}u(\sigma)=I$. For $\sigma>1$ we may write $$u(\sigma)=\frac{1}{2\pi}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\int_{-\infty}^\infty(b/a)^{it}\frac{dt}{(\sigma-1)^2+t^2}.$$ So that $$u(\sigma)=\frac{1}{2\pi}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\frac{\pi e^{-(\sigma-1)|\log(b/a)|}}{\sigma-1},$$ $$u(\sigma)-\frac{1}{2\zeta(\sigma)^2(\sigma-1)}= \frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\frac{e^{-(\sigma-1)|\log(b/a)|}-1}{\sigma-1},$$ and $$u(\sigma)-\frac{1}{2\zeta(\sigma)^2(\sigma-1)}=- \frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma} |\log(b/a)|\int_0^1 e^{-(\sigma-1)|\log(b/a)|x}\,dx.$$ It is not easy here to justify to take limit for $\sigma\to1^+$ term by term, but if correct we will get $$I=-\frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{ab} |\log(b/a)|=\sum_{n=1}^\infty\frac{f(n)}{n}.$$ With $$f(n):=-\frac12\sum_{ab=n}\mu(a)\mu(b)|\log(b/a)|.$$ The function $f(n)=0$ except if $n=mk^2$ with $|\mu(mk)|=1$. In this case $f(n)=f(m)$. For $m$ squarefree $f(m)$ have the sign of $-\mu(m)$ multiplied by a logarithm of a number with prime divisors dividing $m$. But this number depends of the relative size of the divisors of $m$. For example with $p<q<r$ primes $f(pqr)$ can be equal to $2\log (pqr)$ or $3\log r$ according to $pq>r$ or $pq<r$.

This is not properly an answer, after the comments of Tao it is difficult to give an answer. Only an explanation of my comment above. I still think that my series and the integral are equal.

We can write the integral $I$ in the form $$I=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{\zeta(1+it)\zeta(1-it)}\frac{dt}{t^2}.$$ Hence I consider the function $$u(\sigma)=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{\zeta(\sigma+it)\zeta(\sigma-it)}\frac{dt}{(\sigma-1)^2+t^2},\qquad \sigma>1.$$ I expect to have $\lim_{\sigma\to1^+}u(\sigma)=I$. For $\sigma>1$ we may write $$u(\sigma)=\frac{1}{2\pi}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\int_{-\infty}^\infty(b/a)^{it}\frac{dt}{(\sigma-1)^2+t^2}.$$ So that $$u(\sigma)=\frac{1}{2\pi}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\frac{\pi e^{-(\sigma-1)|\log(b/a)|}}{\sigma-1},$$ $$u(\sigma)-\frac{1}{2\zeta(\sigma)^2(\sigma-1)}= \frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma}\frac{e^{-(\sigma-1)|\log(b/a)|}-1}{\sigma-1},$$ and $$u(\sigma)-\frac{1}{2\zeta(\sigma)^2(\sigma-1)}=- \frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{(ab)^\sigma} |\log(b/a)|\int_0^1 e^{-(\sigma-1)|\log(b/a)|x}\,dx.$$ It is not easy here to justify to take limit for $\sigma\to1^+$ term by term, but if correct we will get $$I=-\frac{1}{2}\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{ab} |\log(b/a)|=\sum_{n=1}^\infty\frac{f(n)}{n}.$$ With $$f(n):=-\frac12\sum_{ab=n}\mu(a)\mu(b)|\log(b/a)|.$$ The function $f(n)=0$ except if $n=mk^2$ with $|\mu(mk)|=1$. In this case $f(n)=f(m)$. For $m$ squarefree $f(m)$ have the sign of $-\mu(m)$ multiplied by a logarithm of a number with prime divisors dividing $m$. But this number depends of the relative size of the divisors of $m$. For example with $p<q<r$ primes $f(pqr)$ can be equal to $2\log (pqr)$ or $3\log r$ according to $pq>r$ or $pq<r$.