Q:

establishing a congruence relationship

Is the following deduction correct?
$$ (a+b)\bmod20=y $$
$$\implies (a+b)\bmod20 = y \bmod20$$
$$\implies (a+b) \equiv y\pmod{20}$$

A:

When viewing $\bmod{n}$ as a relation, we are using the definition that $a \equiv b \pmod{n}\,$ means $n \mid b-a$. In this context, I don't think you would normally see that first statement at all, and if you did, it would just be an awkward way of writing the third statement.
It is also possible to view $\bmod{n}$ as an operation, in which case $x \bmod n$ is defined to be the remainder $r$ when $x$ is divided by $n$.
You seem to be using both of these interpretations simultaneously. If you are trying to prove this, I would use the definitions that you are given. If you are going to consider the step from the first to the second equation valid (as in, not needing further explanation), then I don't see that jumping straight from the first to the third would be a problem.
In other words, I would say that either this deduction is trivial enough not to need justification, or else if you are needing to write a proof for some reason, you should use the definitions of the two uses of $\bmod{n}$ and include more detail.

