Both sides count the number of $(b-1)$-subsets of $\{1,\dots,a+b-1\}$ that include $\{1,\dots,b\}$. Because $b > b-1$, this count is obviously $0$, establishing the RHS. For the LHS, apply inclusion-exclusion, where the $b$ properties to be avoided are that $j$ does not appear for $j \in \{1,\dots,b\}$. More generally, this argument shows that $$\sum_{i=0}^b (-1)^i \binom{b}{i} \binom{a+b-1-i}{k} = 0$$ for $k < b$, and it does not require $b \le a$.

Here's a combinatorial proof that the expression is $0$. Both sides of the identity count the number of $(b-1)$-subsets of $\{1,\dots,a+b-1\}$ that include $\{1,\dots,b\}$. Because $b > b-1$, this count is obviously $0$, establishing the RHS. For the LHS, apply inclusion-exclusion, where the $b$ properties to be avoided are that $j$ does not appear for $j \in \{1,\dots,b\}$. More generally, this argument shows that $$\sum_{i=0}^b (-1)^i \binom{b}{i} \binom{a+b-1-i}{k} = 0$$ for $k < b$, and it does not require $b \le a$.