What does elementary function arithmetic without the induction axiom tell us about the exponential function? We can extend Robinson arithmetic (Q) to include exponentiation, using the axioms

1) $x^0=1$ and 2) $x^{S(n)} = x^n*x$. Wiki page

How much can this extension actually say about exponentiation? Because Robinson arithmetic is not associative or commutative, it doesn't seem like proving $x^{a+b} = x^ax^b$ or $(x^a)^b = x^{ab}$ are possible, both of which seem pretty fundamental when it comes to the exponential function.
In fact, I don't even think that proving $x^1 = x$ is possible:
$x^1 = x^{S(0)}*x = 1*x$, but Robinson arithmetic only proves that $x*1 = x$, not that $1*x = x$. I suppose this could be fixed by altering axiom 2), but that still wouldn't solve the other problems. 
I know that Q is weak, but these extra two axioms seem so weak and so bad at capturing exponentiation that it just doesn't even seem worth discussing. Am I missing something?
 A: Your definition of EFA is missing a crucial component. EFA consists of $Q$, the axioms for exponentiation, and induction for formulas with bounded quantifiers. By contrast, $Q$ contains no induction at all, and $PA$ consists of $Q$ together with full induction.
As soon as we add induction to $Q$ - even a small amount of induction, like induction for bounded-quantifier formulas - we get a much more powerful system. In particular, the basic arithmetic facts which $Q$ curiously can't prove tend to be provable from bounded-quantifier induction. It's also worth noting that bounded-quantifier induction in the language including exponentiation is stronger than bounded-quantifier induction in the language of $Q$ alone; that is, EFA is not conservative over the theory $Q$ + induction for bounded-quantifier formulas in the language of $Q$.

For example, here's how to show that $1*x=x$ using bounded-quantifier induction:
There's a slight language-mismatch here. EFA as described has a symbol "$1$" not present in the language of $Q$, and does not contain the symbol "$S$" which is present in the language of $Q$. So the description in the wiki page is a little bit inexact. But we can always replace "$1$" with "$S(0)$" and replace "$S(x)$" with "$1+x$," so we might as well assume we have both $1$ and $S$ to work with.


*

*We have $x*0=0$ as an axiom in $Q$, so in particular we have $1*0=0$.

*Now suppose $1*x=x$. Then $1*(S(x))=(1*x)+1=x+1=S(x).$ Note that "$x+1=S(x)$" really is the right thing to say here: treating "$1$" as an abbreviation for "$S(0)$," the statement "$x+S(0)=S(x)$" is provable in $Q$.

*The formula $\varphi(x)\equiv$ "$1*x=x$" is a bounded-quantifier formula, so applying bounded-quantifier induction to the above two bullets we get $\forall x\varphi(x)$, and we're done.
