The below is too long for a comment so I'm including it here even though I'm not sure it "answers" the question.
If you think about $(1+x)^{-n}$ as living in the ring of formal power series $\mathbb{Z}[[x]]$, then you can show that $$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ and the identity $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ seems very natural. Here's how...
First expand $(1+x)^{-n} = \bigg(\frac{1}{1-(-x)}\bigg)^n = (1 - x + x^2 - x^3 + \dots)^n$. Now, the coefficient on $x^k$ in that product is simply the number of ways to write $k$ as a sum of $n$ nonnegative numbers. That set of sums is in bijection to the set of diagrams with $k$ stars with $n-1$ bars among them. (For example, suppose $k=9$ and $n=4$. Then, **|*|***|*** corresponds to the sum $9=2+1+3+3$; ****||***|**corresponds to the sum $9 = 4+0+3+2$; ****|***||** corresponds to $9=4+3+0+2$; etc.) In each of these stars-and-bars diagrams we have $n+k-1$ objects, and we choose which ones are the $k$ stars in $\binom{n+k-1}{k}$ many ways. The $(-1)^k$ term comes from the alternating signs, and that proves the sum.
There are several ways to derive the result; I’ll start with the one using the binomial theorem. What you have isn’t quite right: the $x$ term in $1-x$ is negative, so it should be
$$(1-x)^{-n}=\sum_{k\ge 0}\binom{-n}k(-x)^k=\sum_{k\ge 0}\binom{-n}k(-1)^kx^k\;.$$
For $\binom{-n}k$ you need to know the full definition of the binomial coefficient: for all real $x$ and non-negative integers $k$ we define
$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}\;;$$
you can check that this agrees with the more familiar definition when $x$ is a non-negative integer. Now we have
$$\begin{align*}
\binom{-n}k&=\frac{(-n)^{\underline k}}{k!}\\
&=\frac{(-n)(-n-1)(-n-2)\ldots(-n-k+1)}{k!}\\
&=(-1)^k\cdot\frac{n(n+1)(n+2)\ldots(n+k-1)}{k!}\\
&=(-1)^k\cdot\frac{(n+k-1)!}{k!(n-1)!}\\
&=(-1)^k\binom{n+k-1}k\;,
\end{align*}$$
so that
$$(1-x)^{-n}=\sum_{k\ge 0}(-1)^k\binom{n+k-1}k(-1)^kx^k=\sum_{k\ge 0}\binom{n+k-1}kx^k\;,$$
since $(-1)^k(-1)^k=(-1)^{2k}=1$.
Another way is to start from the geometric series
$$\frac1{1-x}=\sum_{k\ge 0}x^k\tag{1}$$
and differentiate repeatedly. If I differentiate $(1-x)^{-1}$ with respect $x$ repeatedly, I get $(1-x)^{-2}$, $2(1-x)^{-3}$, $6(1-x)^{-4}$, $24(1-x)^{-5}$, and in general I have
$$\frac{d^n}{dx^n}(1-x)^{-1}=\frac{n!}{(1-x)^{n+1}}\;;$$
this is easy to prove by induction on $n$.
Differentiating the righthand side of $(1)$ $n$ times with respect to $x$, I get
$$\sum_{k\ge 0}k(k-1)(k-2)\ldots(k-n+1)x^{n-k}\;.$$
Setting $\ell=k-n$, so that $k=\ell+n$, I can rewrite this as
$$\sum_{\ell\ge 0}(\ell+n)(\ell+n-1)\ldots(\ell+1)x^\ell=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;.$$
Putting the two pieces together, we see that
$$\frac{n!}{(1-x)^{n+1}}=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;,$$
or, after dividing by $n!$,
$$\frac1{(1-x)^{n+1}}=\sum_{\ell\ge 0}\binom{\ell+n}\ell x^\ell\;.$$
Now just replace $n$ by $n-1$ throughout to get
$$(1-x)^{-n}=\sum_{\ell\ge 0}\binom{\ell+n-1}\ell x^\ell\;,$$
as desired.
Best Answer
It might be easier to see what is going on if we write out some terms: $$ \begin{align} &\overbrace{\sum_{k=0}^N\binom{N}{k}b^{N-k/2}e^{ak/2}}^{b^{N/2}\left(\sqrt{b}+e^{a/2}\right)^N}+\overbrace{\sum_{k=0}^N(-1)^k\binom{N}{k}b^{N-k/2}e^{ak/2}}^{b^{N/2}\left(\sqrt{b}-e^{a/2}\right)^N}\tag1\\[6pt] &=\phantom{2}\binom{N}{0}b^Ne^0+\binom{N}{1}b^{N-1/2}e^{a/2}+\phantom{2}\binom{N}{2}b^{N-1}e^a+\binom{N}{3}b^{N-3/2}e^{3a/2}+\cdots\tag{2a}\\ &+\phantom{2}\binom{N}{0}b^Ne^0-\binom{N}{1}b^{N-1/2}e^{a/2}+\phantom{2}\binom{N}{2}b^{N-1}e^a-\binom{N}{3}b^{N-3/2}e^{3a/2}+\cdots\tag{2b}\\[6pt] &=2\binom{N}{0}b^Ne^0\phantom{{}+\binom{N}{1}b^{N-1/2}e^{a/2}}+2\binom{N}{2}b^{N-1}e^a\phantom{{}+\binom{N}{3}b^{N-3/2}e^{3a/2}}+\cdots\tag3\\[6pt] &=2\sum_{j=0}^{\lfloor N/2\rfloor}\binom{N}{2j}b^{N-j}e^{aj}\tag4 \end{align} $$ The terms with even $k$ in $(1)$ become the terms with $2j$ in $(4)$ and the term with odd $k$ in $(1)$ cancel each other. Since $k$ goes from $0$ to $N$, $j$ goes from $0$ to $\lfloor N/2\rfloor$.
We can substitute $j\mapsto k$ in $(4)$, if desired.