Do SALC-AOs really belong to their symmetry species?
To show that your two functions do represent the $\mathrm{E}$ irreducible representation, we can approach the problem algebraically.

We start by representing the operators of the $C_{3v}$ point group in the basis of the AOs. The identity element is easy in any basis:
$$E=\pmatrix{1 &amp;0 &amp;0\\0 &amp;1 &amp;0\\0 &amp;0 &amp;1\\}$$

We can see that applying this operation to $e_1$ and $e_2$ will just return these functions, so $\chi(E)=2$. We can also represent this operation in the basis of $a_1$, $e_1$, and $e_2$, which will be important for explaining the remaining operations.

For the $C_3$ rotation, we can notice that this just has the effect of moving each AO to the next site (i.e. $\phi_1\to\phi_2$, $\phi_2\to\phi_3$, and $\phi_3\to\phi_1$), so we can write it in matrix form as 
$$C_3=\pmatrix{0 &amp;1 &amp;0\\0 &amp;0 &amp;1\\1 &amp;0 &amp;0\\}$$

Lets see what happens when we apply this to $e_1$:
$$C_3e_1=\frac{1}{\sqrt{6}}\pmatrix{0 &amp;1 &amp;0\\0 &amp;0 &amp;1\\1 &amp;0 &amp;0\\}\pmatrix{2\\-1\\-1}=\frac{1}{\sqrt{6}}\pmatrix{-1\\-1\\2}=e_1'$$

Writing $e_1'$ in the basis of the original functions, we obtain:
$$e_1'=0a_1+\frac{-1}{2}e_1+\frac{-\sqrt{3}}{4}e_2$$

Doing the same with $e_2$:
$$e_2'=0a_1+\frac{-\sqrt{3}}{4}e_1+\frac{-1}{2}e_2$$

We can then take the trace of this matrix (or, to simplify, the 2x2 submatrix spanned by $e_1$ and $e_2$) to get the character of $C_3$:
$$\chi(C_3)=\mathrm{Tr}\pmatrix{\frac{-1}{2} &amp; \frac{-\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} &amp; \frac{-1}{2}}=-1$$

You can go through the same sort of exercise with $\sigma_v$ (determine form in AO basis, apply to each function, write resulting functions in original function basis, take the trace to get the character).