Is there a proof that performing an operation on both sides of an equation preserves equality? so I was learning some abstract algebra and group theory, when they went over the proof of the cancellation law
$$
ab = ac\implies a^{-1}(ab) = a^{-1}(ac)\implies (a^{-1}a)b = (a^{-1}a)c\implies eb=ec \implies b=c
$$
But the first step in which you add the additional term seems jarring to me, especially since I felt we were proving ever trivial thing from the ground up. Obviously I'm familiar with middle school pre-algebra, so I know that it is true that if we perform an operation on both sides it preserves equality, but I didn't know how we know this is the case always. Is it an axiom or is it proven?
Here is my attempt at a proof, let me know if I am going in the correct direction. 
Assume via axiom that $x=x$ and if $a=b$ and $b=c$ then $a=c$, and prove that $a=b\implies ka=kb$
We know that $a=b$, we define $x\mid x=a \implies a=x$. Since $a=b$ and $a=x$, then $b=x$ which we can rewrite as $x=x$. Now we perform the operation on both sides $kx=kx$, which is true via our axiom. Then we re-substitute $x=a$ and $x=b$ to get $ka=kb$. Q.E.D
That was my original idea but I don't know if that's watertight. Thank you!
 A: If the sign "$=$" was some relation that we had just defined on strings like "$ab$", then it would be necessary to establish what you ask for. That is not the case, however. The sign "$=$" denotes equality and juxtaposition of letters denotes multiplication, which is a function. And for every function $f$ you have that $x=y$ implies $f(x)=f(y)$. (Or more similar to the situation here: $x=y$ implies $g(z, x)=g(z, y)$.)
A: The preservation of equations when performing operations follows from the very definition of what operation is. That's all!
Namely, if $\,a=b,\,$ and $\,f\,$ is an operation which applies to the given arguments then $\,f(a)=f(b)\,$ or otherwise $\,f\,$ would not be an operation.
The above applies also to operations in several (or even infinitely many) variables because a collection of variables can be interpreted as a single variable. If you have, say, variables $\,a_1\ldots a_n\,$ then we introduce (explicitly or implicitly) a new variable $\,a:=(a_1\ldots a_n),\,$ and now this is the same story.
REMARK:   More generally, we can consider functions -- the preserving equations by functions is again due to the very definition of the notion of function, just as it was in the case of operations (operations map arguments into the same total set of arguments while function may have their ranges not related to the set of arguments).
A: In first-order logic, we have the formal substitution principle:

Let $\phi$ be a propositional formula with a free variable $v$, and let $\Gamma$ be a context.  Also, let $x, y$ be two terms representing values.  Then:
  \begin{align*} \Gamma & \vdash x = y \\
\Gamma & \vdash \phi[v := x] \\
\hline
\Gamma & \vdash \phi[v := y]. \end{align*}

Informally, what this says is: if you can prove in some context that $x=y$, and you can also prove some statement is true for $x$, then you can conclude the same statement is true for $y$.  (The notation $\phi[v := x]$ just means the result of substituting $x$ in for $v$ in the proposition formula $\phi$.)
Now, if we are working in a group, let us apply this to the formula $\phi := (a^{-1} (ab) = a^{-1} v)$.  Then in the context of the proof, we are assuming $ab = ac$.  Also, $\phi[v := ab]$ results in the proposition $a^{-1}(ab) = a^{-1}(ab)$, which is true by the first-order axiom (or in some formulations, the formal proof rule) of reflexivity of equality: $t = t$ for any term $t$.  Therefore, the substitution principle allows us to conclude that $\phi[v := ac]$ is true, which results in $a^{-1} (ab) = a^{-1} (ac)$.
To give another application which is implicitly used in the proof, let us see how to use the substitution principle to prove the transitivity of equality: if we have $x=y$ and $y=z$ then $x=z$.  For this proof, we will use $\phi := (x = v)$.  Then we are assuming $y=z$.  We also have that $\phi[v := y]$ is true since it reduces to the assumption $x = y$.  Therefore, we can conclude $\phi[v := z]$ which is just $x = z$.
A: We usually take it as an axiom of the relation of equality. In particular, we assume that on a set $E$ there is a relation $=$ which satisfies the following properties for $a,b,c\in E:$
(1) $a=a$ for all $a\in E$ (egoism),
(2) If $a=b,$ then $b=a$ (reciprocity),
(3) If $a=b$ and $b=c,$ then $a=c$ (continuity),
(4) For any function $f$ defined on $E,$ we have that if $a=b,$ then $f(a)=f(b)$ (conservation).
A: Your proof seems fine, but a bit more complicated than necessary.
We start with x = x by the reflexive property of equality.  Then we substitute each x with ka yielding ka = ka, since the reflexive property of equality holds for all x, and closure holds for any operation by definition.
Then, since a = b, we can choose to only replace the rightmost "a" with "b" and that yields,
ka = kb.
Since we've assumed an arbitrary binary operation here (not a group) which we suppressed writing, for any binary operation F, we have that if a = b, then F(k, a) = F(k, b).
The result has many corollaries, all assuming that x = y, such as F(a, k) = F(b, k), and for a unary operation U, U(x) = U(y), and for any trinary operation T, T(k, a, j) = T(k, b, j).  The most general of which seems as follows:
Theoerm: Suppose that x = y, and that a$_1$, ..., a$_n$ is a complete list L of variables and constants in a formula with an operation F.  Suppose for some a$_k$ in L, a$_k$ = x.  (Note k could equal 1, or k could equal n).  Then 
F(a$_1$, ...,  a$_k$, ..., a$_n$) = F(a$_1$, ..., y, ..., a$_n$).
Proof: I'll leave this as an exercise.  See above for a hint.
