Is this correct?
No.
i) This isn't a confidence interval you're calculating (since those are for parameters or functions of them), nor is it really a prediction interval, a tolerance interval, or any of the more common statistical intervals ... since for starters it's based on known population values, not on a sample.
ii) You already calculated the limits of an interval that includes 95% of the probability; it's $(a,b)$, not $(\mu-a,\mu+b)$.
do we mean that 95% of the values lie within the mean of the random variable
No. The mean is a single value. How can 95% of a continuous distribution lie "within" a single value?
But now we don't have that an approximate 95% confidence interval is [mean−2∗SD,mean+2∗SD] since the pdf of Y is not symmetric.
Just because the density isn't symmetric doesn't of itself mean that a symmetric interval can't include 95% of the probability.
It doesn't include 95%, as it happens, though it's often fairly close to 95% for unimodal distributions. However, while it works pretty well for $\pm 2\sigma$, that doesn't always carry over nearly as well to other numbers of sds not close to 2.
So, is the mean±SD property for a confidence interval only valid for normal random variables?
(Again, keeping in mind that it's not a confidence interval)
Well, actually, for normal random variables, 95% of the distribution is within 1.96 sd's of the mean and 95.4% is within 2 sd's of the mean.
Those numbers are calculated from the normal distribution function; $\Phi(1.96)-\Phi(-1.96)=0.9500$ and $\Phi(2)-\Phi(-2)=0.9545$.
Best Answer
The multiple of a lognormal variable is also lognormally distributed.
$$X \sim \mathcal{N}(\mu,\sigma^2)$$ $$Y = e^X$$ $$CY = Ce^X$$
The random variable $CY$ is lognormally distributed since:
$$\ln(CY) = \ln(Ce^X)=\ln(C)+\ln(e^X)=\ln(C)+X$$
Note that:
$$\ln(C) + X \sim \mathcal{N}(\mu + \ln(C), \sigma^2)$$
Therefore $CY$ is lognormally distributed $CY \sim \ln \mathcal{N}(\mu + \ln(C), \sigma^2)$