The notation is deceptive. $P^A$ is not something constructed from objects $P$ and $A$, but rather something analogous to $P$. In fact, $P$ is a special case of $P^A$, namely $P=P^\varnothing$. The same holds, mutatis mutandis, for $NP$. Removing the contrapositive for extra clarity, the proper way of stating your question is thus:
Why $P^\varnothing=NP^\varnothing$ does not imply $P^A=NP^A$ for every $A$?
Then it should be clear that there is no reason for this to hold, just like, say, $x^0=y^0$ does not imply $x^a=y^a$ for real $x,y,a$.
EDIT: I'm not sure I should mention this, as it will probably just add to the confusion. However, under a proper notion of relativization, the implication $C_1^A\ne C_2^A\Rightarrow C_1\ne C_2$ does actually hold for “small classes” $C_1,C_2$, such as $AC^0[m]$, $TC^0$, $NC^1$, $L$, $NL$. See this paper by Aehlig, Cook, and Nguyen. The main reason which makes it work is that for all these classes, the depth of dependence of the oracle queries on each other is constant, hence any relativized function can be written as a finite composition of unrelativized functions and parallel oracle calls.