Is ZFC without Axiom of Infinity consistent? The incompleteness theorem states that one cannot prove whether ZF or ZFC is consistent, but what about ZFC withouth Axiom of infinity? (Assuming the empty set exists)
Furthermore, let $M$ be a consistent model not invoking infinity and $A,B$ be statements invoking infinity such that $A$ contradicts $B$. Then, let's assume both $M+A$ and $M+B$ are consistent. If statements $\phi_A$ and $\phi_B$ invoking infinity are provable in $M+A$ and $M+B$ relatively, then are finite pieces of $\phi_A$ and $\phi_B$ both provable in $M$?
 A: ZF without the Axiom of Infinity is still strong enough to derive Peano's axioms PA. So we cannot prove its consistency without proving the consistency of PA.
A: The corollary from the incompleteness theorems is that you cannot prove the consistency of $\sf ZFC$ from $\sf ZFC$ itself. You have to have a stronger theory.
For example, in $\sf ZFC+\text{There exists an inaccessible cardinal}$, you can in fact prove the consistency of $\sf ZFC$ because this is a stronger theory.
Similarly this is the case of $\sf ZF_{fin}$ ($\sf ZF$ without infinity). The theory itself cannot prove its own consistency. However $\sf ZFC$ is a strictly stronger theory, and it proves the consistency of $\sf ZF_{fin}$. It does so by exhibiting a set which is a model of the theory, $V_\omega$ - the set of the hereditarily finite sets.
Large cardinal axioms are often called "strong infinity axioms" because they mimic the axiom of infinity, in the sense that they make a stronger theory by describing that a certain set of ordinals exists.
A: Usally "ZFC without Axiom of Infinity" is considered consistent, since $\sf ZFC \setminus  AOI + \neg AOI$ has the natural numbers as a model via Ackermann Coding. If $\sf ZFC \setminus AOI + \neg AOI$ has a model, so does $\sf ZFC \setminus AOI$. This is a model like the Hereditary Finite Sets:

Hereditarily finite set
https://en.wikipedia.org/wiki/Hereditarily_finite_set

The Ackermann Coding was shown here:

Ackermann, Wilhelm (1937), "Die Widerspruchsfreiheit
  der allgemeinen Mengenlehre", Mathematische Annalen,
  114 (1): 305–315

On the other hand, to formally prove it, "ZFC with Axiom of Infinity, the usual ZFC" should do, it would show that $V_\omega$ is a model of $\sf ZFC \setminus AOI + \neg AOI$. So you would need to first accept $\sf ZFC$.
Remark: For the rest of your question concerning $M+A$, I have no answer. How do you combine a model $M$ and a formula $A$?
