What is the probability of a random natural number being a power of $10$

To answer a question such as "what is the probability that ...", you need a precise probabilistic framework to be defined first. When no framework is defined, it's implicitly understood that the uniform distribution is considered.

When it comes to natural numbers, the problem is there is no uniform discrete probability measure there. So the question is ill-defined: what is a random natural number?

Your attempt is the best I can imagine to try and make sense of the question being asked, but it still is a different question.

Edit: @d.k.o.'s answer also comes close to addressing your problem, but you should be very careful that it's not a probability measure in the usual sense, since it's finitely additive instead of countably additive. You should keep that in mind at all times if you wish to remember and cite that example.

You may answer this question by considering a "uniform" distribution over natural numbers that is a charge (finitely additive probability measure) on $(\mathbb{N},2^{\mathbb{N}})$. Specifically, for $A\subset \mathbb{N}$ let $(\delta_A)_n=\mathbf{1}_A(n)$ and let $\mathsf{P}(A):=T(\delta_A)$, where $T$ is a Banach limit on $\ell^{\infty}$.

For example, $\mathsf{P}(\mathbb{N})=1$, $\mathsf{P}(A)=0$ for any finite set $A$, and $\mathsf{P}(\{n\in \mathbb{N}:n\equiv1 \mod 2\})=1/2$.

Now, using your argument, $\mathsf{P}(\{n\in\mathbb{N}:n\text{ is a power of $10$}\})=0$.

What you are computing is not a probability. You can define a uniform probability law on $[1,n]$, compute probability of the event $A:$ $'k\in [1,n] \textit{ is a power of 10 }'$ as a function of $n$, P_A(n), and take limit $$P_A=\underset{n \rightarrow \infty}{lim}P_A(n)=0$$ This is OK, but the limit $P_A$ is not a probability.