Wilsonian and continuum renormalization are distinct ideas.
In Wilsonian renormalisation, we have a theory with a lagrangian, $\mathcal{L}$, defined up to an momentum scale $\Lambda$ with couplings $ \{g_0(\Lambda) \} $. We integrate out modes between a scale $ p \in [s\Lambda , \Lambda]$ for $s<1$. This, in general, causes a shift in the value of the couplings $g_0 \rightarrow g = g_0 + \delta g$. In our final step we rescale momentum and space to get a theory defined up to the same momentum cutoff as before. This way we can directly compare the path integrals of the original and effective theories simply by looking directly at the actions. In general after this rescaling we would find $g \rightarrow s^D (g_0 + \delta g)$ for some value of D.
We use this to roughly classify the couplings into relevant, irrelevant or marginal couplings. Which is mainly determined by the scaling $s^D$.
Now if we go to the continuum renormalization (as for QFT) the procedure is a little different. We have a theory defined at some scale $\Lambda_0$ and we consider what the lower energy effective theory, at a scale $\Lambda$, looks like as this is ultimately the theory that we should see in low energy experiments. This, again, is done by integrating out modes, causing a general shift in couplings $g = g_0 + \delta g$. Now, for continuum renormalization, we dont rescale but rather we ask can we take $\Lambda_0 \rightarrow \infty $ whilst keeping the low energy physics at $\Lambda$ the same. That is, does it have a continuum limit?
Given the absence of a rescaling step in the procedure of continuum renormalization how can we classify couplings by their 'scaling dimenson' when it doesnt even come into play? We might say for example the mass term in a 4D field theory is relevant because it has a scale dimension of $[m^2]=2$.
