Let me first make some comments on the two examples you gave.
The second example is not a known "trick" one would learn in high school. It is a calculus trick, if you can call it that, because it allows you to show that $a'b'$ is close to $ab$ if $a'$ is close to $a$ and $b'$ is close to $b$. So it's really something that would be devised specifically for that purpose, and one wouldn't typically see anything like that before, say, proving that $(x,y) \mapsto xy$ is a continuous function or proving the rule about the derivative of a product. I certainly haven't memorized this identity, but I know from experience (with calculus) that something along those lines is possible, and that's enough.
The first example is the special case of the identity
$$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + y^{n-1})$$
when $n = 3$, $x = a$ and $y = -b$. The general form of this identity is so well-known that if you wanted a list of high school algebra identities similar in importance to that one, the list might be five or ten items long at most. In that case, the other answer's suggestion of Algebra by Gelfand could be what you want.
However, if you want to develop a high level of skill in the manipulative aspects of algebra, there are a couple of sources I can recommend that are at an advanced high school level.
A Problem Book in Algebra, by Krechmar.
Higher Algebra: A Sequel to Higher Algebra for Schools, by Ferrar.
These books were formerly used by students preparing for the most competitive university entrance examinations in the USSR and the UK, respectively. The first book consists of problems with solutions, while the second is an actual textbook.
Edit: Here are two derivations of the second identity that might make it seem more natural. Assume $a'$ is close to $a$ and $b'$ is close to $b$. You are trying to prove $a'b'$ is close to $ab$, but there is no simple way to write the difference between them.
Method 1. Instead, write that $a'b'$ is close to $a'b$, which is in turn close to $ab$:
$$a'b' - a'b = a'(b'-b)$$
$$a'b - ab = (a' - a)b$$
Adding these two equalities, we get
$$a'b'-ab = a'(b' - b) + (a' - a)b.$$
The form that you have is obtained by replacing $a'(b' - b)$ with $a(b' - b)$, which is close to it, plus a correction term.
Method 2. Write $a' = a + \Delta a$ and $b' = b + \Delta b$. Then
$$a'b' - ab = (a + \Delta a)(b + \Delta b) - ab = a \Delta b + b \Delta a + \Delta a \Delta b,$$
which is your identity.