I'll use the $D_6=\langle f,r\mid f^2,r^3,frfr\rangle$ definition for $D_6$. I'm trying to study the structure of $\mathbb Z[D_6]$.
Units
One thing I've noticed $(1+fr)^2=1+2fr+(fr)^2=2(1+fr)$, meaning $1+fr$ is non-invertible.
I'm pretty sure the group of units is all multiples of units from $D_6$ and $\mathbb Z$, i.e. $$\{1,-1,r,-r,r^2,-r^2,f,-f,fr,-fr,fr^2,-fr^2\}$$ although I haven't managed to prove it.
Ideals
I'm pretty sure that all the ideals of $\mathbb Z[D_6]$ is the same as the ideals of $\mathbb Z$, but I've not managed to prove it.
Can you hint at a solution or give one?
Regarding (two-sided) ideals, look for homomorphisms to other rings.
Here is an example. Take the quotient homomorphism $\mathbb{Z}[D_6]\to\mathbb{Z}_5[D_6]$ (i.e., reduction modulo 5); of course the kernel contains the principal ideal generated by 5. Now the ring $\mathbb{Z}_5[D_6]$ is semi-simple because $5\nmid|D_6|=6$; notice that it is also an algebra over the field $\mathbb{F}_5=\mathbb{Z}_5$. This means in must decompose into a product of factors of the form $M_n(\mathbb{F}_{5^s})$ (of dimension $sn^2$) by two theorems of Wedderburn. Now of course the sum of these numbers has to be 6 so there aren't too many choices: in fact we actually get $\mathbb{Z}_5[D_6]\cong\mathbb{F}_5\times\mathbb{F}_5\times M_2(\mathbb{F}_5)$. These arise because there are two 1-dimension representations of $D_6$ over $\mathbb{F}_5$ (the trivial one and the one coming from the quotient homomorphism $D_6\to D_6/\langle r\rangle\cong\{1,-1\}$); there is also a two dimensional matrix representation coming from the fact that $D_6$ is the symmetry group of an equilateral triangle. The decomposition is realised by finding central idempotents which can be written down using characters.
Now projection onto each of the three factors composed with the original quotient homomorphism gives three homomorphisms with maximal ideals as their kernels. In fact these are the only maximal ideals in $\mathbb{Z}[D_6]$ that contain 5.
If you replace 5 by either of the primes 2 or 3 things are not as easy because the Jacobson radical is not trivial and $\mathbb{F}_p[D_6]$ is not semi-simple, but you can kill it and then decompose the quotient which is semi-simple, but the dimension will be less that 6 in those cases. I think you still get $M_2(\mathbb{F}_p)$ occurring in each case.