Evaluating $\int_{0}^{1}\frac{x-1}{(x+1)\ln x} dx $
We know that $$\int_0^1 x^y \, \mathrm{d}y=\frac{x-1}{\ln{x}}$$ Therefore, \begin{align*} \int_{0}^{1}\frac{\color{red}{x-1}}{(x+1)\color{red}{\ln x}} \mathrm{d}x &= \int_0^1 \color{red}{\int_0^1} \frac{\color{red}{x^y}}{1+x} \, \color{red}{\mathrm{d}y} \, \mathrm{d}x\\ &=\int_0^1 \int_0^1 \sum_{n=0}^{\infty} {(-1)}^n x^{n+y} \, \mathrm{d}x \, \mathrm{d}y\\ &= \sum_{n=0}^{\infty} {(-1)}^n \int_0^1 \int_0^1 x^{n+y} \, \mathrm{d}x \, \mathrm{d}y \tag{1}\\ &= \sum_{n=0}^{\infty} {(-1)}^n \int_0^1 \frac{1}{y+n+1} \, \mathrm{d}y \\ &= \sum_{n=0}^{\infty} {(-1)}^n \left(\ln{(n+2)}-\ln{(n+1)}\right) \\ &= \ln{\left(\prod_{n=0}^{\infty} \frac{{(2n+2)}^2}{(2n+1)(2n+3)}\right)} \tag{2}\\ &= \boxed{\ln{\left(\frac{\pi}{2}\right)}} \\ \end{align*}
$(1)$: See Sangchul Lee's comment
$(2)$: Wallis product
Here, we will more focus on answering to OP's specific questions:
1. The substitution $x\mapsto1/x$ yields
$$ I = \int_{1}^{\infty} \frac{x-1}{x^2(x+1)\log x} \, \mathrm{d}x. $$
So it seems that OP made a mistake when applying the substitution.
2. Continuing from OP's approach, the recurrence relation and $J(\infty)=0$ together imply
$$ J(a) = \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots,$$
and hence
\begin{align*} -I'(a) &= -\frac{1}{a+1} + 2 \biggl( \frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+3}-\dots \biggr) \\ &= \biggl( \frac{1}{a+1} - \frac{2}{a+2} + \frac{1}{a+3} \biggr) + \biggl( \frac{1}{a+3} - \frac{2}{a+4} + \frac{1}{a+5} \biggr) + \dots \\ &= \sum_{n=1}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr). \end{align*}
Now we integrate both sides from $0$ to $\infty$. Then the left-hand side becomes $I(0)$ by $I(\infty) = 0$. On the other hand, each term of the summation is non-negative, so we can apply the Fubini-Tonelli Theorem to interchange the order of integration and summation to get
\begin{align*} I(0) &= \int_{0}^{\infty} (-I'(a)) \, \mathrm{d}a \\ &= \sum_{n=1}^{\infty} \int_{0}^{\infty} \biggl( \frac{1}{a+2n-1} - \frac{2}{a+2n} + \frac{1}{a+2n+1} \biggr) \, \mathrm{d}a \\ &= \sum_{n=1}^{\infty} (-\log(2n-1) + 2 \log(2n) - \log(2n+1)) \\ &= \log \Biggl( \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \Biggr). \end{align*}
By using the Wallis product formula, the product term reduces to $\frac{\pi}{2}$, proving
$$ I(0) = \log\left(\frac{\pi}{2}\right). $$