If $y$ is an exact solution to the ODE $y'=f(x,y)$ and $y_i=y(x_i)$ for each $i$, where $x_{i+1}=x_i+h$, then Taylor's formula implies $ y_{i+1}=y_i+hf(x_i,y_i)+o(h)\qquad\text{as $h\to0$}. $ (If you're not familiar with little-oh and its big brother big-oh notation, see the wikipedia article on it.) For reasonably smooth functions $f$, the $o(h)$ term can be replaced by $O(h^2)$.
If now $(y_i)$ comes from a numerical method, you say the method is consistent if the same formula holds. Think of it as a minimal requirement for the method to give approximations to the exact solution: Without consistency, the method is no good at all.
Along with a consistency proof you often get better estimates on the $o(h)$ term that can yield a convergence proof.
In your case, just knowing that $k_2=f(x_i,y_i)+O(h)$ (where I assume that $f$ is smooth) gives you $ y_{i+1}=y_i+h\cdot(a+b)f(x_i,y_y)+O(h^2), $ and so the method is consistent if and only if $a+b=1$.