This is how it can be done "by hand", that is without charachter theory. All of this can be found (in more compressed form) at the end of first lecture in Fulton, Harris "Representation theory: the first course". I'll write an outline.
First we classify the irreducible representatins of $S_3$: if $\tau = (123)$, $\sigma = (12)$, then $\tau$ has an eigenvector $v$ with an eigenvalue $\omega$ which is one of the three cubic roots of 1. Then $\sigma(v)$ is also an eigenvector with an eigenvalue $\omega^2$. This follows from $\sigma\tau\sigma = \tau^2$. Now, if $\omega\ne 1$, the representation is irreducible. This shows that an irreducible representation of $S_3$ has degree at most 2. Also, every two-dimensional irreducible representation of $S_3$ is isomorphic to this. Now, if the representation is one-dimensional, then it's generator is an eigenvalue of $\sigma$ and since the order of $\sigma$ is 2, it's eigenvalues are $\pm 1$. Therefore, there are two one-dimensional representations - the trivial one and the one for which $\sigma(v) = -v$. The last one is easily seen to be the sign representation.
To summarise: two one-deminsional representations and one two-dimensional one (up to isomorphism). Also, any two-dimensional representation is generated by eigenvectors $\{x,y\}$ of $\tau$. If the corresponding eigenvalues are 1, then it is decomposable into one-dimensional subrepresentations generated either by $x$ and $y$ (if \sigma(x)\in, $\sigma(y) = y$) or by $x+\sigma(x)$ and $x-\sigma(x)$ (if $\sigma$ permutes ,). In both cases one of these is the trivial and the other one is the sign representation.
It is left to notice that when acting on the regular representation (dimension 6) $\tau$ cyclicly permutes two sets of basis vectors as follows: $ (12)\mapsto(13)\mapsto(23)\mapsto(12) $ and $ Id\mapsto\tau\mapsto\tau^2\mapsto Id. $ This means that the matrix of $\tau$ in the basis $\{(12),(13),(23),Id,(123) = \tau, (132) = \tau^2\}$ has a block-diagonal form (easy to write out, but tideous to type) and the charachteristic polynomial of $\tau$ is $(t^3 - 1)^2$. This shows that $\tau$ has every cubic root of 1 as an eigenvalue and the multiplicity of each is 2. Combine this with the fact that $\tau$ is diagonalisable and the reasoning above, and we get that the regular representation decomposes into three 2-dimensional ones, 2 of which are siomorphic irreducible and the third itself decomposes into the trivial and the sign.