If you are looking for a solution of an equality $a = x\mod{b}$, instead of what we usually care about, i.e. the congruence, you have to be clear about what $x \mod{b}$ means. I also wonder what kind of solution you are looking for? Are you looking for integers or natural numbers?
I think what you are intending is that $x \mod{b}$ is the unique integer $n$ satisfying $0\le n < b$ and $b | x-n$, in other words the remainder of $x$ after division by $b$.
From this it's clear that your equation does not always have a solution. If $a\ge b$ it is impossible to solve the equation.
However if the given numbers satisfy $a<b$, you can find infinitely many solutions.
Then what you write for $x = nb+a$ is close to correct, the only thing is you have to allow for $n$ to be zero, which I don't think is included in your concept of natural numbers.
To write it out clearly:
$
\begin{eqnarray}
&a = x \mod {b}\\
&\Leftrightarrow 0 \leq a < b \text{ and } b|x-a
\end{eqnarray}
$
At this point we distinguish two cases one where $a \ge b$ and we don't have solutions, and one where $a<b$ and we do have solutions.
In this case the divisibility condition is equivalent to stating
$
\exists n\in\mathbb{Z}: bn = x-a.
$
So all the solutions are of the form $bn + a$ for $n\in \mathbb{Z}$. If you are interested in integer solutions for $x$, then you are done at this point. If instead you are looking for natural numbers you need to solve $bn+a> 0$.
I would suggest that instead of taking this approach you become more comfortable with congruences. What I've done above is far less useful than solving the congruence $a \equiv x\mod{b}.$ Congruences are not at all sloppy, or "abusive", they are very rigorously defined. It's a good first example of a finite group and a finite ring, both are fundamental concepts in mathematics.
