"Does this mean implication is superfluous shorthand?"
No. Implication might come as a pre-assumed notion. It can qualify as a superfluous shorthand in a system where the meaning of implication is not pre-assumed. But, that all depends on the system at hand.
"So long rules of weakening and modus ponens are part of the logic, it seems easy to prove A⊢B from ⊢A→B."
Yes, that's true. Also, you can prove it just from modus ponens.
"Is it therefore only in the other direction that the deduction theorems can fail (if A⊢B then ⊢A→B)?"
No. Logical systems for classical propositional calculus without modus ponens do exist. John Halleck's page indicates that Jean Porte described such a system a while back. If modus ponens fails, you don't have the first direction.
"When the deduction theorem does fail, where exactly does it fail in the proof? The base step or the inductive step?"
It can fail in either step. I'll note that your source on The Deduction Theorem uses an axiomatic context. Suppose we have some system with modus ponens still in place.
First, suppose that the well-formed formula (A -> (B -> A)) does not qualify as provable in the system. Then, the part of the meta-proof for the base case does not go through. This is the case in relevance logic.
Second, suppose that the well-formed formula ((A -> (B -> C)) -> ((A -> B) -> (A -> C))) does not qualify as provable in the system, but (A -> (B -> A)) does qualify as provable. Then, the base case will go through, but then case iv. of the inductive step does not go through. This is the case in 3-valued Lukasiewicz logic, and in infinite-valued Lukasiewicz logic.
Best Answer
In the common standard usage $A \vdash B$ means that there is a proof in the relevant deductive system of $B$ from the premiss $A$. And the deduction theorem says that if $A \vdash B$ then $\vdash A \Rightarrow B$, where the double arrow is the conditional in the deductive system.
You can't write $(A \vdash B) \vdash (A \Rightarrow B)$ since what is on the left of second turnstile is not standardly a wff of the object language but an abbreviatory symbol added to the metalanguage and so a fortiori there is no formal proof in the system from a premiss of the form $A \vdash B$.
You could write $(A \vdash B) \to\ \vdash (A \Rightarrow B)$, where the single arrow is a conditional in the metalanguage, and the double arrow is the object language conditional. But you can't substitute the turnstile for the single arrow or fuse the two.
(If the turnstile belongs to a sequent calculus, parallel remarks still apply.)