I am looking to calculate
$$\sum^{n}_{j = 0}{{a + j}\choose{b}}$$
where $a, b \in \mathbb{Z}$. I realize that if $b = a$ then this is proven here with a different labeling of variables, but in my case $a$ and $b$ are different.
Among the things I have tried that did not work were the Snake Oil method and applying Hockey-Stick Identities (such as this one).
Either an analytic or combinatorial proof is fine.
On
This is not the most direct approach but it does show where hypergeometric functions appear as mentioned by @DrSonnhardGraubner.
We write $$ \begin{align} \sum_{j=0}^n\binom{a+j}{b} &=\sum_{j=0}^n\frac{\Gamma(a+1+j)}{\Gamma(b+1)\Gamma(a-b+1+j)}\\ &=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j}{(a-b+1)_j}\\ &=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\sum_{j=0}^n\frac{(a+1)_j(1)_j}{(a-b+1)_j\,j!}, \end{align} $$ where $(s)_n=\Gamma(s+n)/\Gamma(s)$ is the Pochhammer symbol and $(1)_n=n!$. The result is a partial hypergeometric series, which may be evaluated using DLMF 16.2.4 to give $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}\frac{(a+1)_n(1)_n}{(a-b+1)_n\,n!}{_3F_2}\left({-n,1,b-a-n\atop -n,-a-n};1\right). $$ After some simplification $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}{_2F_1}\left({1,b-a-n\atop -a-n};1\right). $$ Gauss's hypergeometric theorem then permits us to write $$ \sum_{j=0}^n\binom{a+j}{b}=\frac{\Gamma(a+1+n)}{\Gamma(b+1)\Gamma(a-b+1+n)}\frac{\Gamma(-a-n)\Gamma(-1-b)}{\Gamma(-a-n-1)\Gamma(b)}, $$ which can be further reduced to give a sum of two binomial coefficients.
Hints: Notice that if $a<b=a+c$ then it is the same thing as the case $a=b$ because $$\sum _{j=0}^n\binom{a+j}{b}=\sum _{j=-c}^{n-c}\binom{a+c+i}{b}=\sum _{j=0}^{n-c}\binom{b+i}{b},$$ can you conclude?
For the case in which $b<a$ then you can write the sum as suggested by peterwhy in the comments i.e.,
$$\sum _{j=0}^{a-1}\binom{j}{b}+\sum _{j=0}^{n}\binom{a+j}{b}=\sum _{j=0}^{a+n}\binom{j}{b},$$ can you conclude?