Determine filter type (HPF or LPF) by its coefficients
I am not sure if my answer will correctly suits your question as it seems a little confusing. Anyway, I understand that you want to eliminate the initial transition of your filter. What I propose is to initialize your filter state so that the output equals the input at $t=0$

I had to solve that problem in the past. What I did was to assume that all delays were initialized to the same value. Doing that, you simplify your problem a lot. I don't have a lot of theory to back this technique, but that seems to work quite good on my side.

Let's start with Direct Form II



If we consider that all delays are set to the same value $K$, we find this output equation.

$$ y = (-a_1k-a_2k+x)b_0+b_1k+b_2k $$
A little algebra : 
$$ y = b_0k(-a_1-a_2-...)+b_0x+k(b_1+b_2+...)$$
$$ y = -kb_0\sum_{n=1}^{N_a}{a_n} + b_0x+k\sum_{n=1}^{N_b}{b_n}$$

Now, since we know we want the output to be equal to the input, we can say $y=x$

$$ x = -kb_0\sum_{n=1}^{N_a}{a_n} + b_0x+k\sum_{n=1}^{N_b}{b_n}$$

Then

$$ x-b_0x = -kb_0\sum_{n=1}^{N_a}{a_n}+k\sum_{n=1}^{N_b}{b_n}$$
$$ x(1-b_0) = -kb_0\sum_{n=1}^{N_a}{a_n}+k\sum_{n=1}^{N_b}{b_n}$$
$$ \frac{x(1-b_0)}{-b_0\sum_{n=1}^{N_a}{a_n}+\sum_{n=1}^{N_b}{b_n}} = k $$

So there you go. Initialize all your delays with $K$ and your filter will start with an output equal to the input. 

You can do this for other forms as well.

Hope that helps