I know that by their respective definitions:
$$\gamma_0 = \sum_{n=1}^{\infty}{\frac{1}{n}-\ln(\frac{n+1}{n})}$$
$$\ln(n)=\sum_{d|n}{\Lambda(d)}$$
and I want to get there:
$$\gamma_0 = \sum_{n=1}^{\infty}{\frac{1-\Lambda(n)}{2n}}$$
I suspect that a number theoretic aproximation must be made regarding to the sum over divisors of $n$ and divisors of $n+1$, but no idea from where can it come.
As others have noted, it's hard to prove more than$$\lim_{s\to1^+}\sum_{n\ge1}\frac{1-\Lambda(n)}{n^s}=\gamma_0,$$which I'll focus on. (In particular, partial summation of $\sum_{n\ge1}\frac{1-\Lambda(n)}{n}=\gamma_0$ implies the prime number theorem viz. $\sum_{n=1}^N(\Lambda(n)-1)\in o(N)$.) In terms of Dirichlet convolutions $\ln=\Lambda\ast1\implies\Lambda=\ln\ast\mu$. We want to evaluate at $s=1$ the Dirichlet series of $\tfrac{1-\Lambda}{2}$. Since $1,\,\ln,\,\mu$ have respective Dirichlet series $\zeta,\,-\zeta^\prime,\,1/\zeta$, we want$$\lim_{s\to1^+}\frac{\zeta(s)+\zeta^\prime(s)/\zeta(s)}{2}=\gamma_0.$$Now use the $0<s-1\ll1$ results$$\begin{align}\zeta(s)&\in-\tfrac{1}{1-s}+\gamma_0+\gamma_1(1-s)+O((1-s)^2),\\\zeta^\prime(s)&\in-\tfrac{1}{(1-s)^2}-\gamma_1+O(1-s),\\\tfrac{\zeta^\prime(s)}{\zeta(s)}&\in\tfrac{1}{1-s}+\gamma_0+O(1-s),\\\zeta(s)+\tfrac{\zeta^\prime(s)}{\zeta(s)}&\in2\gamma_0+O(1-s).\end{align}$$