Is this sufficient condition for infinite number generation from a finite seed also a necessary one?

Given an initial finite set of natural numbers (the seed). Use the following rule to add numbers to the set. If $a,b,a+b \in S \rightarrow ab \in S$. Where $S$ is the is set. For example take the initial seed $\{2,3,5\}$.

$2,3,(2+3) \in S \rightarrow (2*3) \in S \quad$ So now the new set is $\{2,3,5,6\}$

$3,3,(3+3) \in S \rightarrow (3*3) \in S \quad$ So now the new set is $\{2,3,5,6,9\}$

$3,6,(3+6) \in S \rightarrow (3*6) \in S \quad$ So now the new set is $\{2,3,5,6,9,18\}$

$9,9,(9+9) \in S \rightarrow (9*9) \in S \quad$ So now the new set is $\{2,3,5,6,9,18,81\}$

This seven-element set ends the generation, no new numbers can be added.

Now let's look at a second example where the initial seed is $\{2,3,4,5\}$

$2,3,(2+3) \in S \rightarrow (2*3) \in S \quad$ So now the new set is $\{2,3,4,5,6\}$

$2,4,(2+4) \in S \rightarrow (2*4) \in S \quad$ So now the new set is $\{2,3,4,5,6,8\}$

There is enough at this point to show that the set will be infinite.

Let $x \in \Bbb{N}\space|\space x\gt 1$. If there is the pattern of elements $x,2x,3x,4x$ in $S$ then the rule will produce infinite numbers. (In this example we have $2,4,6,8$.)

$x,x,(x+x) \in S \rightarrow (x*x) \in S$

$x,2x,(x+2x) \in S \rightarrow (x*2x) \in S$

$x,3x,(x+3x) \in S \rightarrow (x*3x) \in S$

$2x,2x,(2x+2x) \in S \rightarrow (2x*2x) \in S$

This means the new set will now contain $x^2, 2x^2, 3x^2, 4x^2$. This is inductive.

This condition of having $x,2x,3x,4x$ in the set is sufficient to show that there is an infinite number of numbers that will be generated.

My question is this a necessary condition to produce an infinite set? In other words does there exist a finite seed that can generate infinite sets without any subset (from the seed or generated sets) of the form ${x,2x,3x,4x}$?

(@Ingix showed that the previous version of the question wasn't explicit enough the new formulation was made because of his critique)

One point that I want to make explicitly here is $S\subset\Bbb{N}$ at all times. This avoids the issue of having a set like $\{2,3,5,5,5,5,5,5,...,5\}$. Where you could technically have an infinite set but not address the problem I'm trying to solve.

Technically you can find seeds not following the pattern. For example, for $x>2$ start with

$$S=\{2,x,x+2,4x,6x,8x\}.$$

Of course, you immediately conclude that $2,x,x+2 \in S \Rightarrow 2x \in S$, which produces the "forbidden" pattern in the form of $2x,4x,6x,8x$. It should also be clear that you can go back this way as far as you want, say setting $x=ab$ and start with $a,b,a+b$ instead of $x$, a.s.o.

So a better question would likely be: Do there exist infinte sets generated from finite seeds without any subset of the form $\{x,2x,3x,4x\}$? Which I unfortunately don't know the answer to.