This is exactly what I was looking for. Thanks!

@Carl, This is indeed a nice example. I can see exactly what goes wrong. PA encodes all Turing machines. Knowing a completion of PA solves the halting problem for each encoded machine (I think!). But computing the completion relative to an oracle doesn't give the encoded Turning machines access to this oracle. If, however, we added a function to the language of PA and hard coded the oracle into the function by adding axioms, then any completion of PA with these new axioms is not computable relative to the oracle. This is making me believe relativization is more of an art.

@qiaochu - I should clarify myself. First, despite it being my only example, I'm not as interested in complexity theory as I am in the type of computability theory done in math departments, computable analysis, various degrees of computability, and algorithmic randomness. (Although there are plenty of CS people who also work on this.) Most proofs I see in these fields are clearly relativizable. It's often mentioned as an aside. I was looking for a simple or well known proof that isn't and where it fails. I hope this helps.

OK, I have it for the finite case at least. Your problem can be reduced to a system of linear equations with one variable for each node $a$ representing the value $f(a)$. Then, if $L$ is finite, so is the system, and we can pick a (possibly empty) set of variables, with corresponding nodes $a_1,\ldots,a_n$, that are free and the rest are determined from them. Hence if $M= \\{ a_1,\ldots,a_n \\}$, then any $f:M\rightarrow \mathbb{R}$ extends uniquely to an additive function $f$ on $L$. I don't know if this property is true of infinite systems of finite linear equations.

Actually, my above comment is incorrect as well. Knowing $a,0$ determines $b,c,d,1$. We must have $a=b=c=d$, and then $0+1=2a$. So maybe there is something salvageable about my argument.

@Hendrik - Yes, I seemed to have neglected the possibility of over-specification. Worst, consider the lattice $0,1,a,b,c,d$ where $a,b,c,d$ are incompatible and $0,1$ are the smallest, largest elements. This has no subset $M$ on which every $f:M \rightarrow R$ extends uniquely to an additive $f$ on the lattice.