It is not exactly equal to $x$ except when $x=0$. What was likely meant is that if $|x|$ is not far from $0$, then $-\log(1-x)$ is approximately equal to $x$.
One formal way of putting it is that $\lim_{x\to 0} \frac{-\log(1-x)}{x}=1.$
Take for example $x=0.1$. The calculator gives that $-\log(0.9)\approx 0.1053605$, so for $x=0.1$, $-\ln(1-x)$ is indeed fairly close to $x$. The second term of the Taylor Series tells you roughly how big the error is when you approximate $-\log(1-x)$ by $x$. Note that if $x=0.1$, then $x^2/2=0.005$. The actual error is approximately $0.0053605$.
Sometimes, in modelling physical situations, when we know that $|x|$ is close to $0$, we replace $-\ln(1-x)$ by $x$. That does not mean that the two are strictly equal, only that for our purposes the approximation is good enough.
The equation of the tangent line to $y=-\log(1-x)$ at $x=0$ turns out to be $y=x$. Recall that the tangent line to $y=f(x)$ at $x=a$ kisses the curve $y=f(x)$ at $x=a$. So if $x$ is near $0$, then $-\log(1-x)$ is very close to $x$. We call $x$ the linear approximation to $y=-\log(1-x)$ near $x=0$. Similarly, $x+\frac{x^2}{2}$ is the quadratic approximation.
Added: Your edit shows some uncertainty about the inequalities $x\leq \int_{1-x}^{1} \frac{dt}{t} \leq \dfrac{1}{1-x} x.$ That uncertainty is reasonable, particularly for negative $x$. For geometric clarity we should treat the cases x>0 and x<0 separately.
First assume that $x$ is positive.
On our interval from $1-x$ to $1$, the function $\dfrac{1}{t}$ is always $\ge 1$. So the integral (area) is bigger than or equal to $1$ times the length of the interval, which is $x$. That proves the inequality on the left. (It is good to draw a picture.)
Similarly, on our interval, $\dfrac{1}{t}$ is always less than or equal to $\dfrac{1}{1-x}$. So the integral is less than or equal to $\dfrac{1}{1-x}$ times the length of the interval. That yields the inequality on the right.
Next assume that $x$ is negative. It is all too easy to make mistakes with negative numbers, so we temporarily let $w=-x$. Then $w$ is positive.
We are looking at the integral from $1-x$ to $1$, so from $1+w$ to $1$. This is is going the wrong way. But $\int_{1+w}^1 \frac{dt}{t}=-\int_1^{1+w}\frac{dt}{t}.$ By an argument close to the one given above for the case $x>0$, we find that $\frac{w}{1+w}\le \int_1^{1+w}\frac{dt}{t}\le w.$ "Multiply" the inequality by $-1$. That reverses the inequality, and we obtain $-w\le \int_1^{1+w}-\frac{dt}{t}\le -\frac{w}{1+w}.$ Finally, replace $w$ by $-x$, and interchange the bounds on the integral. We get $x\leq \int_{1-x}^{1} \frac{dt}{t} \leq \frac{x}{1-x}.$