I think that is not correct (maybe I'm mistaken, but in any case is great to see other options).
You want to calculate the Shannon entropy of finding a prime. You suppose that the probability of finding a number in a sequence of lenght is $1/n$, so the information you have of the system is $\log(n)$, so the information per number is $\log(n)/n$. Until that point I agree with you.
But now you say that if you have $m$ primes in the sequence then every number gives $m \log(n) / n$ per prime, and say that the total information about them is 1 because of the prime number theorem...
Not sure about that. Let's say you have $m$ primes in your sequence. Then, if I pick randomly a number, the probability to obtain a prime is $m/n\simeq 1/\log(n)$, so the information associated with this is $ S=\log_2(\log(n)) / \log(n)$. Note that if $n \rightarrow +\infty$ this tends to 0. This is because the probability to get a prime number in the limit tends to 0 (if it's nearly impossible to get a prime number, then you have a lot of certainty that the number will not be prime and the information you get is very small). This makes a lot more sense than a constant value, given that as you make $n$ larger is more and more difficult to find primes.
Your reasoning is good, but I think that you should not sum the entropies of every prime number, because you're triying to determine the entropy associated with "finding a prime number", and what you do is summing every entropy.
Ah, and another funny thing. In information theory usually $\log(x)$ stands for $\log_2(x)$ because the information is measured in bits. That implies that, even if your reasoning is correct, the Shannon entropy (in bits) is $\log(n)/\log_2(n) = \log(2) \simeq 0.69$ bits. The result you gave is in nats.