There is a better abstract-algebra proof, but I will here give somewhat simpler version.
Let $X$ be an additive inverse of $x$, that is $x + X = 0$. Note that equation $X = 0 - x = - x$ holds $\forall x, X \in \text{G}$ where $G$ is some group ($0$ is additive identity).
We can subtract $X$ from both sides: $x + X - X = 0 - X.$ Thus, $x = - (-x)$. In simpler terms, subtracting $- x$ is adding $x$.
There are plenty of geometric answers there so I will not 'steal' and write them here. I will just mention that semantic $+ (+ x) = +x$ holds because of covention (easier is to manipulate symbols like that). That pluses esentially aren't the same thing, they are different operations. Simillary for $- \ x \equiv 0 - x$ and $-x$ which is additive inverse. Because of that we have $- \ (-x) \neq -x$ or $- \ x$ for $x \neq 0$.
You will easly see that geometrical argument for $\pm a \cdot \pm b = \pm c$ (of course, signs can be in different order!) and also take for simplicity $a, b, c \in \mathbb{N}$, is built on convention that first $+$ is right (then first $-$ is left) and second $+$ is adding to the previous orientation (then second $-$ is subtracting from previous orientation which is same as adding to inverse orientation of the previous one), third $+$ is position of resulting number, right from $0$, then we write $-$ if result is left from $0$.
Also see this:
$0 - 2 = 2', 0 - 1 = 1', 0 - 0 = 0.$ You see that results are increasing, so $0 - 1' = 1.$
