Forgive me in advanced if this is a trivial question.
This convention makes perfect sense to me intuitively, but is there any rigorous underpinning to it?
I'm beginning to read through an abstract algebra textbook, and soon after establishing what a ring is, it all of the sudden added to both sides of the equation to prove a(0)=(0)a=0. It seemed a little premature to me.
It comes down to what the equals sign means. It means, as @MaliceVidrine says, that the two things mentioned on either side of the equation are the same thing. If you do something to the left side, you have presumably changed it. If you don't do the same thing to the right side, you are now asserting that two different things are the same, and you have told a lie.
If $ a = b $ in a set $ S $ and $ f: S \to T $ is a function, then $ f (a)=f (b) $ (this is the definition of a function). Doing the same thing to both sides of an equation in, say, a ring, is a special case of this. (e.g. there is a set function $ R \to R $ given by $r \mapsto r + 3$ and adding 3 to both sides of an equation is applying this function.
This may sound a little daft, but I seldom see this issue explained in what I think is the simplest way: you have to apply any operation to both sides of an equation because it's the same entity being referred to on both sides. For some anonymous function $f$ and $x,y$ such that $x=y$, to apply $f$ to $x$ and not to $y$ is to apply $f$ to $x$ and also not apply $f$ to $x$ (since $x=y$).
You can, in practice, actually end up with $f(x)=y\wedge x=y$, but what this means is that $f(x)=x$ (that is, $x$ is a fixed point of $f$). If we were able to pass from an equality to a one-sided application of a function as a general rule, we would be accepting as a general principle that every point is a fixed point of every function.
Here's a short example of doing the same thing to both sides of an equation, and a justification for each step, using only substitution and the field axioms. The magic happens in the first two steps.
Statement:
\begin{align*} a+c=b+c \Rightarrow a=b \end{align*}
Proof:
\begin{align*} (a+c)+(-c)&=(a+c)+(-c) \tag{property of =} \\ (a+c)+(-c)&=(b+c)+(-c) \tag{substitution}\\ a+(c+(-c))&=b+(c+(-c)) \tag{associativity}\\ a+0&=b+0 \tag{inverses}\\ a&=b \tag{identity} \end{align*}
$\square$
Consider addition as a function in two variables, like $f(x,y) := x+y$. If you have an identity such as $x_1 = x_2$ then you have $f(x_1, y) = f(x_2, y)$, because basically you insert the same arguments into the function. So if you do this, then you still have the equality $x_1+y = x_2+y$ (by definition of f). By the same reasoning, you can of course replace $y$ by $y_1$ and $y_2$ and you can justify the equivalent statement for other operations such as ring multiplication.
