Counting the ways on grid if one can move from $(x,y)$ to $(x+a, x+b)$ for arbitrary $x,y,a,b\geq 0$. On 2 dimensional grid, consider the situation that one can move from $(p,q)$ to $(p+α,q+β)$ at once for arbitrary integer $p,q,α,β\geq 0 \land (α,β)\neq(0,0)$.
I want to count how many ways are there to move from (0,0) to (x,y).
I proved there is $\sum_{i=0}^{\min(x,y)}\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}$ by combinatorial view. 
Then, can we derive this using formal power series?
I've tried to derive this, however different formula appear and I cannot get the combinatorial interpretation of that formula.
The number of ways to get $(x,y)$ by $n$ moves is
\begin{align}
&[s^x t^y]\left(\frac{1}{1-s}\frac{1}{1-t}-1 \right)^n \\
=&[s^x t^y]\left(\frac{s+t-st}{(1-s)(1-t)}\right)^n
\end{align}
Note that $[s^x t^y] f(s,t)$ is the coefficient of $s^x t^y$ term of $f(s,t)$.
Summing up for $n=1,2,...,$ we can get the number of ways to go to $(x, y)$ by arbitrary number of moves. 
\begin{align}
&[s^x t^y]\sum_{n=1}^\infty \left(\frac{s+t-st}{(1-s)(1-t)}\right)^n \\
=&[s^x t^y]\frac{s+t-st}{1-2(s+t-st)} \\
=&[s^x t^y]\sum_{i=0}^{\min(x,y)}2^{x+y-i-1} (s+t-st)^{x+y-i} \\
=&\sum_{i=0}^{\min(x,y)}2^{x+y-i-1} (-1)^i \frac{(x+y-i)!}{(x-i)!(y-i)!i!}
\end{align}
However, this seems different from $\sum\binom{x}{i}\binom{y}{i}2^{x+y-(i+1)}$.
Also, I cannot come up with the combinatorial interpretation of the formula we get.
UPDATE
I want to explain in details for the following.
\begin{align}
&[s^x t^y]\sum_{n=1}^\infty \left(\frac{s+t-st}{(1-s)(1-t)}\right)^n \\
=&[s^x t^y]\left(\frac{s+t-st}{1-2(s+t-st)}  - \frac{(1-s)(1-t)\lim_{N\to\infty}\left(\frac{s+t-st}{(1-s)(1-t)}\right)^N}{1-2(s+t-st)} \right)\\
\end{align}
Here, I suppose the term, $-\frac{(1-s)(1-t)\lim_{N\to\infty}\left(\frac{s+t-st}{(1-s)(1-t)}\right)^N}{1-2(s+t-st)}$ can be treated as $0$ because if we put $s=0$ and $t=0$, $\frac{s+t-st}{(1-s)(1-t)}=0$ which means the degree of this term will go $\infty$ if we take power of $\infty$. Thus this term have nothing to do with the $s^x t^y$ term and it's ok to treat it as $0$.
 A: We consider non-negative integers $x,y$ and to  get  a first  impression we start calculating the first few values of
\begin{align*}
\sum_{j\geq 0}\binom{x}{j}\binom{y}{j}2^{x+y-{j+1}}\tag{1}
\end{align*}
We write $j\geq 0$ and recall $\binom{p}{q}=0$ if $q>p$. The values of (1)  are given in the picture below and we observe the sequence is archived in OEIS as A059576.
                                          
The values in OEIS coincide  with (1)  besides $(x,y)=(0,0)$  which  is  set  to $1$, so  that the  value of $(x,y)$ is the sum  of the values  with smaller  $x$ or smaller  $y$ (an example marked in blue).

We now assume $x,y\geq  0, x+y\geq 1$ and obtain
  \begin{align*}
\color{blue}{[s^xt^y]}&\color{blue}{\sum_{n=1}^\infty\left(\frac{s+t-st}{(1-s)(1-t)}\right)^n}\\
&=[s^xt^y]\left(\frac{1}{1-\frac{s+t-st}{(1-s)(1-t)}}-1\right)\\
&=[s^xt^y]\frac{s+t-st}{1-2(s+t-st)}\\
&=\frac{1}{2}[s^xt^y]\frac{1}{1-2(s+t-st)}\tag{2}\\
&=\frac{1}{2}[s^xt^y]\sum_{j=0}^\infty 2^j(s+t-st)^j\\
&=\frac{1}{2}[s^xt^y]\sum_{j=0}^\infty2^j \sum_{k=0}^j\binom{j}{k}s^k(1-t)^kt^{j-k}\\
&=\frac{1}{2}[s^xt^y]\sum_{k=0}^\infty\sum_{j=k}^\infty  2^j\binom{j}{k}s^k(1-t)^kt^{j-k}\tag{3}\\
&=\frac{1}{2}[t^y]\sum_{j=x}^\infty 2^j\binom{j}{x}(1-t)^xt^{j-x}\tag{4}\\
&=\frac{1}{2}[t^y]\sum_{j=0}^\infty 2^{j+x}\binom{x+j}{j}t^j(1-t)^x\\
&=\frac{1}{2}\sum_{j=0}^y2^{j+x}\binom{x+j}{j}[t^{y-j}](1-t)^x\\
&=\frac{1}{2}\sum_{j=0}^y2^{j+x}\binom{x+j}{j}\binom{x}{y-j}(-1)^{y-j}\tag{5}\\
&=\sum_{j=0}^y\binom{x+y-j}{y-j}\binom{x}{j}2^{x+y-j-1}(-1)^{y-j}\tag{6}\\
&=2^{x+y-1}\sum_{j\geq  0}\binom{x}{j}\left(-\frac{1}{2}\right)^j[z^{y-j}](1+z)^{x+y-j}\\
&=2^{x+y-1}[z^y](1+z)^{x+y}\sum_{j\geq 0}\binom{x}{j}\left(-\frac{z}{2(1+z)}\right)^j\\
&=2^{x+y-1}[z^y](1+z)^{x+y}\left(1-\frac{z}{2(1+z)}\right)^x\\
&=2^{x+y-1}[z^y](1+z)^{y}\left(1+\frac{z}{2}\right)^x\\
&=2^{x+y-1}[z^y](1+z)^{y}\sum_{j\geq 0}\binom{x}{j}\left(\frac{z}{2}\right)^j\\
&=\sum_{j\geq 0}\binom{x}{j}[z^{y-j}](1+z)^y2^{x+y-j-1}\\
&=\sum_{j\geq 0}\binom{x}{j}\binom{y}{y-j}2^{x+y-j-1}\\
&\,\,\color{blue}{=\sum_{j\geq 0}\binom{x}{j}\binom{y}{j}2^{x+y-j-1}}
\end{align*}
  and  the claim follows.

Comment:


*

*In (2) we use $\frac{2(s+t-st)}{1-2(s+t-st)}=\frac{1}{1-2(s+t-st)}-1$. We can ignore the term $1$ which does not contribute to $[s^xt^y]$ since $x+y\geq 1$.

*In (3) we exchange the summation of series.

*In (4) we  select the coefficient  of $s^x$.

*In (5) we select the coefficient of $t^{y-j}$.

*In (6) we change the order of summation $j\to y-j$.
Note: The expression with the exponent $\infty$ is mathematically not sound and should be avoided.
