Aren't the non-negativity constraints special cases of the general constraints?
Isn't $$x_1 \ge 0$$ just $$-1 \cdot x_1 + 0 \cdot x_2 \le 0?$$
Then you could just summarize the general form of linear programming as $$\max\{\mathbf{c}^T\mathbf{x} | A\mathbf{x}\le \mathbf{b} \}.$$
Or is it that the non-negativity constraints are a must for most LP algorithms to work?
It's due to the fact that by having them, we ensure that the feasible region of linear model is bounded in some fashion, and this allows optimization algorithms to, like the Simplex algorithm (which is an algorithm that depends on non-negative constraint of some form), find extreme points on these bounds. However, they are entirely optional as better explained by Prubin in his answer to the question "Why does the simplex algorithm not accept negative decision variables?" located here. In general though, they're like any other constraint, and are often excluded in the general abstraction like how you observed.
To explain the optionality of these constraints for the general model, in that they're really only important for a linear algorithms, let's consider a model that has only free variables with the in the perspective of the Simplex method. Suppose we have the following linear model:
$$\max z= 5x + 5y$$ $$\text{Subject to: }\qquad\qquad\qquad$$ $$x + y \le 9$$ $$3x + y \le 18$$ $$(x,y)\in\mathbb{R}^2$$
Which will have the following feasible region with only one extreme point:
Since the Simplex algorithm is one that requires non-negativity constraints for it's decision variables, we can use substitution methods to "add non-negativity" to the previous model like so:
Let $x = x' - x''$ with $x'\ge 0$ and $x''\ge0$.
Let $y = y' - y''$ with $y'\ge 0$ and $y''\ge0$.
Substituting the above into the previous model we get:
$$\max z= 5x'-5x'' + 5y'-5y''$$ $$\text{Subject to: }\qquad\qquad\qquad$$ $$x' -x'' + y'-y'' \le 9$$ $$3x'-3x'' + y'-y'' \le 18$$ $$x',x'',y',y''\ge0$$
Which can be solved with the Simplex algorithm. Thus, while they're not really special, they are needed in some form for linear solvers to derive a solution, which can be shown in both models here and here.