Let $A(n,m)$ denote a set such that:
$$A(n,m) = \{k\in\mathbb{N}_{>0} \mid (j \in \mathbb{N}) \wedge (k \neq jm)\} \implies$$ $$A(n,2) = \{1,3,5,7,\dots\}$$ $$A(n,4) = \{1,2,3,5,6,7,9,\dots\}$$
To be more explicit: $A(n,m)$ denotes the $n$th natural number larger than $0$, which is not a product of $m$ and any natural number, $j$.
How could one prove that:
$$A(n,m) = \left \lfloor \frac{n m -1}{m-1} \right \rfloor$$
Any help with tagging more appropriately would be appreciated!
To be clear/ensure I've understood your definition, you have defined a set $$ S(m) = \{k \in \mathbb{N} \mid k \not\equiv 0 \pmod{m}\}, $$ and when viewed as an increasing sequence $S(m) = \{s_{m,i}\}_{i=1}^\infty$ (where $s_{m,i} \leq s_{m,i+1}$ for all $i$), your function $A(m,n)$ is simply $$ A(m,n) = s_{m,n}. $$
If my interpretation of the problem is correct (as it seems to be), then this follows from the division algorithm.
Note that the only natural $k$ that are omitted from $S(m)$ are the multiples of $m$, i.e., $$ S(m) = \bigcup_{q \in \mathbb{N}\cup\{0\}}\{qm+r \mid r=1,2,\ldots,m-1\}. \qquad (\star) $$ Fix $m,n \in \mathbb{N}$, and assume $m > 1$. (Note that your claim is false if $m=1$.) Then $n = q(m-1) + r$ for $q = \lfloor n/(m-1) \rfloor \geq 0$ and $0 \leq r < m-1$. (The reason for the choice of divisor $m-1$ instead of $m$ is because each of the sets in the union in $(\star)$ is of order $m-1$.)
It's not hard to see from either the definition of $s_{m,n}$ or $(\star)$ that $$ s_{m,n} = qm+r, $$ and so it is enough to show that the right hand side of this equation equals $\lfloor nm/(m-1)\rfloor$. This can be derived as follows:
\begin{align*} \left\lfloor\frac{nm}{m-1}\right\rfloor &= \left\lfloor \frac{(q(m-1)+r)m}{m-1} \right\rfloor\\ &= \left\lfloor qm + \frac{rm}{m-1} \right\rfloor\\ &= qm + \left\lfloor r + \frac{r}{m-1} \right\rfloor\\ &= qm + r = \left\lfloor \frac{n}{m-1} \right\rfloor m + r \end{align*}