Must a Set's size be a natural number, and would the contrary invalidate a solution?

What you have discovered is that the formula $2^n - 2^{n-2}$ has a continuous extension from the natural numbers $n \in \{1,2,3,...\}$ to all real numbers $x \in \mathbb{R}$; in other words, the formula $2^x - 2^{x-2}$ makes sense no matter what the value of $x$.

This is a significant discovery. It is not at all obvious, from an elementary (precalculus) point of view, that an exponential function like $2^n$ can be extended in a continuous fashion to a function $2^x$ so that the laws of exponents remain valid, laws such as $2^x \cdot 2^y = 2^{x+y}$ and so on. It requires a semester or two of calculus to prove this rigorously, by a method which rigorously constructs the function $2^x$ and proves its properties.

However, you should not over-interpret this discovery. You are right that it makes no sense to ask how many subsets there are in the set with $9.58$ elements, so that is not what your formula says.

The reason this works is that the two conditions (i): at least one from $\{1,2\}$, and (ii): at least two from $\{3,4,5\}$, are independent when $n\geq5$. The probability that (i) is satisfied is $3/2^2$, and the probability that (ii) is satisfied is $4/2^3$. It follows that ${3\over8}$ of all $2^n$ subsets satisfy both conditions. Taking logarithms at the second step was a detour of yours that Shannon might have liked $\ldots$.