I believe I have figured out this proof, though I'm hoping someone can look it over for style and notation. In particular, in modular arithmetic, I find myself going back and forth between equivalence, $\equiv$, and a standard equals sign. I believe the distinction is that the latter follows from some kind of algebra, the former from equivalence in a modulus. I've been stringing these together in equation chains, though, which I'm unsure is valid. Here's the theorem and my proof.
Theorem: If $x$ and $y$ are invertible in $\mathbb{Z}_m$, then $xy$ and $x^{-1}$ are also invertible.
Proof: If $x$ and $y$ are invertible, $\exists a, b$ s.t. $xa \equiv \ \text{(mod $m$)}$ and $yb \equiv 1 \ \text{(mod $m$)}$. Then, \begin{align*} \left(xy\right)\left(ab\right) & = \left(xy\right)\left(ba\right) \\ & = x \left(yb\right) a \\ & \equiv x \cdot 1 \cdot a \ \text{(mod $m$)} \\ & = xa \\ & \equiv 1 \ \text{(mod $m$)} \end{align*} and, thus, $ab$ is the inverse of $xy$.
$x^{-1}$ is trivially invertible, as $xa = xx^{-1} = x^{-1} x \equiv 1 \ \text{(mod $m$)}$, with an inverse of $x$.
So, my questions are: (1) How do these proofs look? (2) Is the equality chain valid? In other words, can I conclude that $yb \equiv 1 \ \text{(mod $m$)}$, and then continue the chain with, say, algebraic simplification and then introduce another equivalence? This might be trivial, though I'm not too familiar with the conventions of modular arithmetic. It seems that some people write these with a standard equality sign but indicate which modulus they're working in.
Thanks in advance.