Relation between edit distances over different alphabets
Obviously, $L(2^b) \leq L(2)$.

The Levenshtein distance is at most the length of the longest string, so $L(2^b) \leq \max(cn,cm) / b$ whereas $L(2) \leq \max(cn,cm)$.

You cannot avoid $L(2^b)(S_1,S_2) = L(2)(S_1,S_2)$ when $L(2)(S_1,S_2) \leq \max(cn,cm) / b$:
consider for instance $b = c = 2$, $S_1 = 0^{2n}$ and $S_2 = (01)^{p} (00)^{n-p}$.

When $L(2)(S_1,S_2) &gt; \max(cn,cm) / b$, you cannot avoid $L(2^b)(S_1,S_2) = \max(cn,cm) / b$.
Consider for instance $b = c = 2$, $S_1 = 0^{2n}$ and $S_2 = 1^{2n}$.