Proof that Mellin transform of random variable determines distribution

Question

For $X$ a nonnegative real random variable, define the Mellin transform of the distribution $\mathbb P_X$ to be the function $m_X : [0,\infty) \to [0,\infty]$ given by $$ m_X(s) = \mathbb E[X^s] $$ I'm trying to prove the following:

Theorem. Assume there is an $\epsilon_0 > 0$ with $m_X(\epsilon_0) < \infty$. Then for any $\epsilon > 0$, the distribution $\mathbb P_X$ is characterized by the values $m_X(s)$ over $s \in [0,\epsilon]$.

My reference text suggests proving this result in the following steps:

1. Conclude the statement for a random variable $X$ with a continuous density.
2. For $\sigma > 0$, let $Y_\sigma$ be uniformly distributed over $[1-\sigma, 1]$ and be independent of $X$. Show that $XY_\sigma$ has a continuous density.
3. Compute $m_{XY_\sigma}$, and show that $m_{XY_\sigma} \to m_X$ for $\sigma \searrow 0$.
4. Show that $XY_\sigma$ converges in distribution to $X$ as $\sigma\searrow 0$.

Step 1 is easy using the inverse Mellin transform law for continuous functions. Step 2 I'm having some trouble with. Actually I don't think it's true as stated; suppose $X = 1$ almost surely. Then $\mathbb P_{XY_\sigma} = \mathbb P_{Y_\sigma}$ is the normalized Lebesgue measure on $[1-\sigma, 1]$, which does not have a continuous density with respect to Lebesgue measure on $[0,\infty)$.

Question 1. For which $Y_\sigma$ independent of $X$ and for which $Y_\sigma$ converges in distribution to $1$ do we have that $XY_\sigma$ has a continuous density function?

I tried showing Step 2 assuming instead that the distribution $Y_\sigma$ has a continuous density given by $f_\sigma$ and that the distribution $\mathbb P_{Y_\sigma}$ converges weakly to the Dirac measure at $1$. Even then, I'm having trouble showing the density would be continuous: by independence of $X$ and $Y_\sigma$, the density of $XY_\sigma$ would be given by \begin{align*} \frac{d}{dx} \mathbb P[XY_{\sigma} \leq x] = \frac{d}{dx} \int_{\mathbb R} \mathbb P[X \leq x/y | Y_\sigma = y] f_\sigma(y) \, dy = \frac{d}{dx} \int_0^\infty \mathbb P[X \leq x/y] f_\sigma(y) \, dy \end{align*} but without knowing if the CDF of $X$ is differentiable, I'm not convinced we can easily compute this derivative.

Assuming we could show $XY_\sigma$ has a continuous density, Steps 3 and 4 should be straightforward with something like dominated convergence and the Portemanteau theorem. But if we know the values of $m_X(s)$, without knowing anything about the continuity of the Mellin transform itself, I'm not seeing how to conclude that $\mathbb P_X$ is determined by $m_X$.

Question 2. Assuming $m_{XY_\sigma} \to m_X$ pointwise as $\sigma \to 0$ and that $XY_{\sigma} \to X$ in distribution, given that the distributions $\mathbb P_{XY_\sigma}$ are determined by the functions $m_{XY_\sigma}$, how do we conclude that $\mathbb P_X$ is determined by $m_X$?

Any ideas? I'd also accept a reference to the Mellin transform for random variables as an answer.