AFAIK it's because SVMs are often used together with kernels.
if we don't use kernels, it is sufficient to only store the decision boundary $wx+b=0$, then SVM will become a parametric method.
kernels can be thought of as mapping the input to an implicit feature space, of which the dimensionality can be very high (even infinite in the case of Gaussian kernels). in this case storing the decision boundary on the high dimensional space would be inefficient (or impossible if there're infinite dimensions).
if we decompose the decision boundary as a function of some support vectors (as shown in Daneel Olivaw's answer) then it would be independent of the dimensionality of the "implicit feature space". such that it can work with any type of kernels.