Trying to determine equivalence of two definitions of multiplication for integers modulo $n$ Consider an integer modulo $n$, defined in terms of the following coset of the integer group with addition operation, for a coset representative $a$:
$$\overline{a} = \{a + k n \, | \, k \in \mathbb{Z}\}$$
We define $\overline{b}$ similarly, where $a, b < n$. Multiplication is defined in this way: $\overline{a} \cdot \overline{b} \equiv \overline{a b}$, where the latter $a b$ employs the usual integer multiplication (I use $\equiv$ to mean "is defined as"). See for example Dummit & Foote, Preliminaries section.
I wanted to check if this is equivalent to the following definition (I don't think it is in general, but am having trouble showing whether it is or it isn't):
$$\overline{a} * \overline{b} \equiv \{xy \, | \, x \in \overline{a}, y \in \overline{b}\}$$
If you write out my second proposed definition explicitly for the set members, you'll see that whether it is equivalent seems to rest on whether the following equation has integer solutions $(k,l)$ for all $N \in \mathbb{Z}$:
$$ a l + b k + n k l = N$$
Whether this has integer solutions for all $N$ determines whether the resulting set is equal to the coset $\overline{a b}$ that results from the first (standard) definition. This looks like a nonlinear diophantine equation or the like, which I haven't dealt with before. I feel like I'm overthinking this or am missing something.
Edit: It was requested that I show how I obtained the equation and why it must have solutions for all $N$. From my definition of $*$, we have:
$$
\begin{eqnarray}
\overline{a}*\overline{b} &\equiv& \{x y \, | x \in \overline{a}, y \in \overline{b}\} \\
&=& \{ (a + kn)(b + l n) \, | \, k,l \in \mathbb{Z} \} \\
&=& \{ ab + (a l + b k + n k l) n \, | \, k,l \in \mathbb{Z} \} \\
\end{eqnarray}
$$
If the definitions are equivalent, the set in the last equality must be the same as $\{ab+Nn \, | \, N \in \mathbb{Z}\}$. For this to be true, $a l + b k + n k l$ must span all the integers just as $N$ does. So, $a l + b k + n k l$ must be surjective on $\mathbb{Z}$, and therefore $a l + b k + n k l = N$ must have integer solutions $(k,l)$ for all integers $N$. Recall that $a$, $b$, and $n$ are all fixed.

No it isn't. Working mod $5$, we get $\overline 2\cdot\overline 2=\overline 4=\{5k+4\mid k\in\mathbb Z\}$. 
So $19\in\overline 4$, but $19\not\in\{xy\mid x,y\in\overline 2\}$, since the only ways to express $19$ as a product are $1\cdot19$ and $(-1)\cdot(-19)$, and none of these factors are in $\overline 2$.
