Proof verification: show that $x - a = b$ can be rewritten as $x = b + a$. I am beginning an introductory college math course to catch up from my bad high school education. This is one my first proofs.

Prove that $x - a = b$ can be rewritten as $x = b + a$.

We have been given the properties of the operations of the set of the real numbers (not sure how to latex that).
My proof is this:
$x - a - (-a) = b - (-a)$
$x - 0 = b + a$
$x = b + a$
I'm not completely sure this is correct.
I have another, more important and general doubt. In the proof I use the fact that adding something to both sides of an equation does not change the equation. Do I need to prove this, since no proof has been given in this course, if I want to use it? We are proving very intuitively obvious theorems, so I'm not sure what other intuitively obvious theorems I can use without proving first!
Here's an attempt:
Theorem: adding $x \in R$ to both sizes of an equation does not change the equation.
If $a, b$ are real numbers and $a = b$, then $a$ and $b$ are the same. $a + x = b + x$ can then be rewritten as $a + x = a + x$, since a = b. Both sides are the same, so $a + x = b + x$.
Here I'm not sure how to say that a bunch of operations in the real numbers is a real number. This also should work for all operations we haven't mentioned yet: if $a^{2/3} = b^{4/7}$, then $a^{2/3} + c = b^{4/7} + c$. I'm not sure if this complicates things, but I have no idea how to say this either way.
Let's also ignore for a second that this is an introductory course. Would I need to prove this if I was asked to prove the theorem in an exam?

As you say, $$x-a=b\implies (x-a)+a=b+a$$ because both members represent the same values before and after addition (actually, $\iff$ holds.)
Then by properties of the addition,
$$(x-a)+a=x+(-a+a)=x+0=x.$$
