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Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is a model ZF including Regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

4. I have to confess that I am not sure what you mean by the Abstraction Scheme. Is this what I would call Separation? Anyhow, given an uncountable regular cardinal $\kappa$ (such as $\aleph_1$),
   $H_\kappa$, the collection of all sets whose transitive closure is of size $<\kappa$, is a model of ZFC without the Power Set Axiom (I am again assuming ZFC as my background theory). It follows that there is no procedure to build a model of ZFC from a model of ZFC without Power Set. So it seems that any minimal subsystem of ZFC would have to include the Power Set Axiom.

Partial Conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, but you do need Infinity and Power Set.

Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is a model ZF including Regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

4. I have to confess that I am not sure what you mean by the Abstraction Scheme. Is this what I would call Separation? Anyhow, given an uncountable regular cardinal $\kappa$ (such as $\aleph_1$),
   $H_\kappa$, the collection of all sets whose transitive closure is of size $<\kappa$, is a model of ZFC without the Power Set Axiom (I am again assuming ZFC as my background theory). It follows that there is no procedure to build a model of ZFC from a model of ZFC without Power Set. So it seems that any minimal subsystem of ZFC would have to include the Power Set Axiom.

Partial conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, you don't need Separation if your version of Replacement is sufficiently strong, but you do need Infinity and Power Set.

Note the we have used exactly two types of arguments here (except for the obvious "Axiom A implies Axiom B, so we don't need B"): To show that from a model of some subsystem of ZFC we can construct a model of ZFC, just give the construction and why it works (constructible sets, well founded sets). To show that some axiom is necessary, prove the existence of a model of ZFC without the axiom in ZFC, i.e., show that ZFC implies Con(ZFC without the axiom). The Second Incompleteness Theorem then tells you that there is no procedure to construct a model of ZFC from a model of ZFC without the axiom.

Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is an inner model that satisfies regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

4. I have to confess that I am not sure what you mean by the Abstraction Scheme. Is this what I would call Separation? Anyhow, given an uncountable regular cardinal $\kappa$ (such as $\aleph_1$),
   $H_\kappa$, the collection of all sets whose transitive closure is of size $<\kappa$, is a model of ZFC without the Power Set Axiom (I am again assuming ZFC as my background theory). It follows that there is no procedure to build a model of ZFC from a model of ZFC without Power Set. So it seems that any minimal subsystem of ZFC would have to include the Power Set Axiom.

Partial Conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, but you do need Infinity and Power Set.

Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is a model ZF including Regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

4. I have to confess that I am not sure what you mean by the Abstraction Scheme. Is this what I would call Separation? Anyhow, given an uncountable regular cardinal $\kappa$ (such as $\aleph_1$),
   $H_\kappa$, the collection of all sets whose transitive closure is of size $<\kappa$, is a model of ZFC without the Power Set Axiom (I am again assuming ZFC as my background theory). It follows that there is no procedure to build a model of ZFC from a model of ZFC without Power Set. So it seems that any minimal subsystem of ZFC would have to include the Power Set Axiom.

Partial Conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, but you do need Infinity and Power Set.

Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is an inner model that satisfies regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

Just some minor points:

1. The situation with the Axiom of Regularity (well-foundedness of the $\in$-relation) is similar to the situation with the Axiom of Choice. In any model of ZF without regularity we can build the usual von Neumann hierarchy $V_\alpha$, $\alpha$ an ordinal, by iterating the power set operation, and the union of the $V_\alpha$'s is an inner model that satisfies regularity.

2. Depending on the precise formulation, the Replacement Scheme often implies the Separation Scheme.

3. As you indicate, Infinity is indispensable, since $V_\omega$ is a model of all the other axioms, and even if the background theory is full ZFC (i.e., if we pretend to live in a universe satisfying ZFC), from $V_\omega$ we cannot build a model of ZFC.

4. I have to confess that I am not sure what you mean by the Abstraction Scheme. Is this what I would call Separation? Anyhow, given an uncountable regular cardinal $\kappa$ (such as $\aleph_1$),
   $H_\kappa$, the collection of all sets whose transitive closure is of size $<\kappa$, is a model of ZFC without the Power Set Axiom (I am again assuming ZFC as my background theory). It follows that there is no procedure to build a model of ZFC from a model of ZFC without Power Set. So it seems that any minimal subsystem of ZFC would have to include the Power Set Axiom.

Partial Conclusion so far: You need infinitely many axioms by Joel's answer, you don't need AC or Regularity, but you do need Infinity and Power Set.