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Say I have a set $S={1, 4, 10, 7}$. Could I then multiply $S$ by $3$? Would that then look like $3S={3, 12, 30, 21}$? Any help would be really appreciated.

Sure, we sometimes for example denote the set of even integers by $2Bbb Z={dots,-4,-2,0,2,4,dots}$, while the set of integers is $Bbb Z={dots,-2,-1,0,1,2,dots}$.

Yes..you have already defined the operation..a scalar multiplication on a set.

In fact, this is closely related to images of functions:

Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := {f(a) | a in A'}quad left(= {b in B | exists a' in A'. f(a') = b}right)$$

Your idea is a specific instance:

Let $A' = {1, 4, 10, 7} subseteq mathbb{N} =: A$ and $f: mathbb{N} to mathbb{N}$ be the operation "multiply by three", then
$$f[A'] = {3, 12, 30, 21}$$

The other answer hints at the notation $NA'$, e.g. $3A'$.

Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbb{Z} = {0}$. However, with this "function framework", it is easy to state a criterion for when the equality holds:

If $f$ is injective on $A'$, then $|A'| = |f[A']|$.

Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."

Can you see why $|2mathbb{Z}| = |mathbb{Z}|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)