Here's a nice example that I think is easier to understand than the usual examples of Goodstein's theorem, Paris-Harrington, etc. Take a countably infinite paint box; this means that it has one color of paint for each positive integer; we can therefore call the colors $C_1, C_2, $ and so on. Take the set of real numbers, and imagine that each real number is painted with one of the colors of paint.
Now ask the question: Are there four real numbers $a,b,c,d$, all painted the same color, and not all zero, such that $$a+b=c+d?$$
It seems reasonable to imagine that the answer depends on how exactly the numbers have been colored. For example, if you were to color every real number with color $C_1$, then obviously there are $a,b,c,d$ satisfying the two desiderata. But one can at least entertain the possibility that if the real numbers were colored in a sufficiently complicated way, there would not be four numbers of the same color with $a+b=c+d$; perhaps a sufficiently clever painter could arrange that for any four numbers with $a+b=c+d$ there would always be at least one of a different color than the rest.
So now you can ask the question: Must such $a,b,c,d$ exist regardless of how cleverly the numbers are actually colored?
And the answer, proved by Erdős in 1943 is: yes, if and only if the continuum hypothesis is false, and is therefore independent of the usual foundational axioms for mathematics.
The result is mentioned in
Fox says that the result I described follows from a more general result of Erdős and Kakutani, that the continuum hypothesis is equivalent to there being a countable coloring of the reals such that each monochromatic subset is linearly independent over $\Bbb Q$, which is proved in:
A proof for the $a+b=c+d$ situation, originally proved by Erdős, is given in:
The sentence $1=0$ is false, so it is (hopefully) unprovable. On the other hand it is disprovable (in any halfway reasonable theory for reasoning about the integers) and therefore decidable.
In order to be undecidable, both the sentence and its negation must be unprovable. In other words, your "a sentence which is unprovable, and also its negation is unprovable" is exactly what "undecidable" means.
Note that this is always relative to a particular theory or proof system. Something can't just be "undecidable" in and of itself; but it can be "undecidable in ZF" or "undecidable in PA", for example
(Beware that in the related area of computability theory, "undecidable" has a completely different meaning, and is synonymous with "non-computable". The two meanings of "undecidable" do not coincide, and don't even apply to the same kind of things. In particular, "computable" doesn't apply to single sentences at all.)
But what about unprovable sentences that are not undecidable? Can they also be added as axioms?
Such sentences are necessarily disprovable (because if they were not, they would be undecidable). So if we try to add them as axioms, we get an inconsistent theory.
Now, do phrases like "In my opinion [unprovable sentence here] is true" make sense?
Yes, if they are taken to be about truth in the "intended model" of whichever language your theory is phrased in, such as the actual integers. Most of us feel intuitively that the integers exist in some Platonic way, independently of any formal systems for reasoning about them, and that all sentences in the language of arithmetic have a definite (though not necessarily known) truth value when applied to the actual integers.
Best Answer
Yes, you can add the negation of any unprovable statement as a new axiom to the theory and still get a consistent theory out of it.
(If the extended theory proved a contradiction, this would directly be a proof of the original "unprovable" statement, which therefore wouldn't be unprovable after all).
If the new axiom happens to be false in the interpretation of the theory you had in mind (such as the actual $\mathbb N$), this interpretation will not be a model any more, of course. The extended theory will have other models that are non-standard models of your original theory.