(1) According to the problem, we have $$a_n = S_n - S_{n-1} = \frac{3}{2}(a_n - a_{n-1}) \quad (n \geq 2)$$ Therefore, we can simplify this to obtain $$a_n = 3a_{n-1}$$ which leads to $$\frac{a_n}{a_{n-1}} = 3 \quad (n \geq 2)$$ This indicates that for $n \geq 2$, the ratio $\frac{a_n}{a_{n-1}}$ is constant and equal to 3. This is the common ratio of the geometric sequence.

Moreover, for the first term, $$S_1 = \frac{3}{2}(a_1 - 1) = a_1$$ Solving for $a_1$ yields $a_1 = 3$. Since we have a common ratio and a first term, the sequence $\{a_n\}$ forms a geometric sequence.

(2) Since we determined that the sequence $\{a_n\}$ is geometric with a common ratio of 3 and first term $a_1 = 3$, we can write $a_n = 3^n$. Then, $b_n$ can be expressed as $$b_n = \log_3 a_n = n$$ which means that $$c_n = a_n b_n = n \cdot 3^n$$ Let's denote the sum of the first $n$ terms of the sequence $\{c_n\}$ by $T_n$. Consider $$T_n = 1 \cdot 3^1 + 2 \cdot 3^2 + \ldots + (n-1) \cdot 3^{n-1} + n \cdot 3^n$$Multiplying the entire sum by 3, we get $$3T_n = 1 \cdot 3^2 + 2 \cdot 3^3 + \ldots + (n-1) \cdot 3^n + n \cdot 3^{n+1}$$ Subtracting these two equations, we obtain $$-2T_n = (3^1 + 3^2 + \ldots + 3^n) - n \cdot 3^{n+1} = \frac{3(1 - 3^n)}{1-3} - n \cdot 3^{n+1}$$ Simplifying further, we have $$T_n = \frac{(2n-1) \cdot 3^{n+1} + 3}{4}$$ Thus, the sum of the first $n$ terms of $\{c_n\}$ is $$\boxed{T_n = \frac{(2n-1) \cdot 3^{n+1} + 3}{4}}$$