Suppose I draw $N1$ samples from distribution $N(\mu_1,\sigma_1^2)$, $N2$ samples from distribution $N(\mu_2,\sigma_2^2)$. These two distributions are independent.
Can the combined sample of $N1+N2$ observations be treated as samples from a single normal distribution $N(\mu,\sigma^2)$?
Where $\mu = \frac{N1}{N1+N2}\mu_1+\frac{N2}{N1+N2}\mu_2$, $\sigma^2=(\frac{N1}{N1+N2})^2\sigma_1^2+(\frac{N2}{N1+N2})^2\sigma_2^2$.
If yes, then it is great; if not, what is its distribution?
In general, this is a mixture distribution, in this specific case a normal or Gaussian mixture; we have a tag gaussian-mixture-distribution.
It is normal if and only if $\mu_1=\mu_2$ and $\sigma_1=\sigma_2$. You may be able to treat it as "normal enough" if the component means and variances are "similar enough", or if one of the components has a much higher weight. This is just another way of saying that under these circumstances it is hard to reliably infer the number and parameters of components, which is a standard problem in mixture modeling.
If you just combine them as you described (called a "mixture"), the resulting mixed sample (of size $n_1+n_2$) will not have any known, predictable distribution. Take the case of $n$ samples from $N(-10,1)$ and $n$ from $N(10,1)$, you will have a sample from a symmetrical bimodal distribution; definitively not normal. If instead you take $n$ from $N(0,100)$ and $N(0,1)$ you have a sample from a symmetrical unimodal distribution, but with a narrow and tall "hump" superimposed on a wide and shallow "hump"; very non normal. If you mix $2n$ values from $N(0,1)$ with $n$ values from $N(2,1)$, your mixed data will now come from a skewed distribution; very non-normal.
Now, your mixed sample will have for a mean the "mixture" of the sample means, as you indicated in the question (for estimation of the mean -first order approximation-, you can replace all the sample values by $n$ values at the mean). But the variance of the mixed sample will not at all look like your formula (mixing $N(-100,1)$ and $N(100,1)$ will obviously have a much larger variance than mixing $N(-1,1)$ and $N(1,1)$). And mixing $N(0,1)$ and $N(0,1)$ will still have a variance of 1, regardless of the mixing ratio...
Bottom line: no, the mixture of 2 normal distributions is not normal, the mean of the mixture will be the mixture of the means, but the variance of the mixture will not be at all the mixture of the variances.
I'm a bit sleep and may be misunderstanding your question, but here is my crack at it until somebody smarter comes along.
I'm not sure how you could treat these as their own pre-defined distribution if you have already noted them as being independent with their own means and variances. Consider for example the two distributions $N(0,1)$ and $N(50,1000)$. Their combined distribution would have to come from a very bimodal distribution or something that varies so widely that these subsets of the "mother" distribution hardly represent where they came from. Of course either scenario is possible, but we have no way to actually prove this is the case if either scenario is a possibility.
