The mean and variance of $S_n$ suggest to consider the normalization
$$T_n=n\cdot(S_n-\theta).$$ Note that $E[T_n]=n\cdot(E[S_n]-\theta)=1$ and $\mathrm{var}(T_n)=n^2\cdot\mathrm{var}(S_n)=1$ hence $T_n$ might converge in distribution. Let us now check that this convergence holds.
Since $S_n\geqslant\theta$ almost surely, $T_n\geqslant0$ almost surely. For every $t\geqslant0$, $[T_n\geqslant t]=[S_n\geqslant\theta+t/n]$ has probability $P[X_1\geqslant\theta+t/n]^n=\mathrm e^{-t}$ hence $T_n$ converges in distribution to a standard exponential distribution.
Note that actually the distribution of $T_n$ does not depend on $n$ since, for every $n$, it is exactly standard exponential.
Verify that for every $\theta>0$, $$E_{\theta}(X_i^2)=\operatorname{Var}_{\theta}(X_i)+[E_{\theta}(X_i)]^2=\frac2{\theta^2}=\mu_\theta\,,\,\text{say}$$
And $$\operatorname{Var}_{\theta}(X_i^2)=E_{\theta}(X_i^4)-[E_{\theta}(X_i^2)]^2=\frac{20}{\theta^4}=\sigma^2_\theta\,,\,\text{say}$$
Taking $Y_i=X_i^2$, you have by classical CLT $$\sqrt n\left(\overline Y-\mu_\theta\right) \stackrel{L}\longrightarrow N\left(0,\sigma^2_\theta\right)$$
Now use Delta-method with the transformation $g(x)=\sqrt\frac{x}{2}$ for $x\ge 0$, so that
$$\sqrt n\left(g(\overline Y)-g(\mu_\theta)\right)\stackrel{L}\longrightarrow N\left(0,[g'(\theta)]^2\sigma^2_{\theta}\right)\tag{*}$$
This $(*)$ would give you the limiting distribution of $g(\overline Y)=\sqrt{\frac{1}{2n}\sum\limits_{i=1}^n X_i^2}$.
Best Answer
The density of the individual $X_i$ is nearly the density of a normal distribution, but it has $\log x$ where one would expect $x$ and that additional $\frac{1}{x}$-factor. That means that the $X_i$ are log-normally distributed, i.e. $\log X_i$ is normally distributed. See http://en.wikipedia.org/wiki/Log-normal_distribution.
By taking the logarithm of the product $S_n = \prod_{i=1}^n X_i$, you get $$ \log S_n = \log \prod_{i=1}^n X_i = \sum_{i=1}^n \log X_i \text{.} $$ $\log S_n$ is thus normally distributed too, and $S_n$ is thus log-normally distributed.
The mean and variable of $\log S_n$ are simply $n$ times the mean respectively variance of $\log X_i$, which can easily be found by looking at the density of $X_i$.
Using that $\log S_n \sim N(\theta n,4n)$, i.e. that $P(\log S_n \leq x) = \Phi\left(\frac{x-n\theta}{2\sqrt{n}}\right)$, where $\Phi$ is the CDF of the standard normal distribution, you get $$ F_n(x) := P(S_n \leq x) = \Phi\left(\frac{\log x-n\theta}{2\sqrt{n}}\right) \text{.} $$ for the CDF of $S_n$.
If $S_n$ is to converge in distribution to $S$ with CDF $F$, then $F_n \to F$ pointwise, except for discontinuity points of $F$.
You thus have the following cases.
If $\theta > 0$, $F_n(x) \to 0$ as $x \to \infty$ for all $x > 0$ and thus by monotonicity any possible limit distribution would have to be identically zero, which is impossible. The product thus doesn't converge in this case.
If $\theta < 0$, $F_n(x) \to 1$ as $x \to \infty$ for all $x > 0$. In other words, $F(x)=1$ if $x > 0$, and by right-continuity of CDFs also $F(0)=1$. On the other hand, $F_n(x) = 0$ if $x < 0$, and thus $$ F(x) = \begin{cases} 0& \text{if $x < 0$} \\1 &\text{if $x \geq 0$}\end{cases} $$
If $\theta = 0$, $F_n(x) \to \frac{1}{2}$ for all $x > 0$. This is also impossible, since you need $F(x) \to 1$ for $x \to \infty$ if $F$ is to be a CDF.
The product $S$ thus converges exactly if $\theta < 0$, and the limit distribution is a point distribution with $P(S=0)=1$, $P(S\neq0)=0$ in the case.
Intuitively, this makes sense. If the means of the logarithms are less than zero, on average multiplying will make the numbers smaller and smaller. Otherwise, the numbers get larger and larger and the product diverges.