Axiomatizing the (positive-based) exponential function

Question

The real-valued function $\langle x,y\rangle\mapsto x^y$ can be defined in multiple ways, and possible definitions vary somewhat, depending on the desired domain of definition.

For instance, if one wants to define it only for all real $x$ and all positive integers $y,$ the recursive definition $x^1=x,x^{y+1}=x^y\cdot x$ does the trick. From there, if one wants it to be defined for all positive real $x$ and any real $y,$ then one might begin by defining the exponential function $\exp(x)$ by the power series $$\exp(x):=1+\sum_{k=1}^\infty\frac{x^k}{k!},$$ defining the natural logarithm to be the inverse of this function, and then defining $x^y=\exp\bigl(y\cdot\ln(x)\bigr).$ If one wants to extend this function from there, one might use properties expected/desired of the function.

I was curious how uniquely expected/desired properties of the function and domain might determine said function and domain, so I've been playing around with it. In particular, I wondered what one might need to assume about a set $E\subseteq\Bbb R^2$ and a function $g:E\to\Bbb R$ so that $g(x,y)=x^y$ for all $\langle x,y\rangle\in E,$ and so that $\langle x,y\rangle\in E$ whenever $x>0.$

I determined the following:

Suppose $E\subseteq\Bbb R^2$ and $g:E\to\Bbb R$ satisfy the following:

1. For all $x,y,z\in\Bbb R,$ if $\langle x,y\rangle,\langle x,z\rangle\in E,$ then $g(x,y+z)=g(x,y)\cdot g(x,z).$
2. For all $x\in\Bbb R,$ if $\langle x,1\rangle\in E,$ then $g(x,1)=x.$
3. If $x\ge0,$ then $\langle x,1\rangle\in E.$
4. For all $x,y\in\Bbb R$ and positive $m\in\Bbb Z,$ if $\langle x,my\rangle\in E,$ then $\langle x,y\rangle\in E$ and $g(x,y)^m=g(x,my).$
5. For all $\langle x,y\rangle\in E,$ if $g(x,y)\ne0,$ then $\langle x,-y\rangle\in E.$
6. $E$ is convex.
7. $g$ is continuous on $E.$

Then $g(x,y)=x^y$ for all $\langle x,y\rangle\in E,$ and $E=\bigl\{\langle x,y\rangle\in\Bbb R^2:(x>0)\vee(x=0\wedge y>0)\bigr\}.$

To outline a proof: One readily verifies that if the conclusion holds, then so do all of the desired hypotheses and conditions. Applying 1, 2, 3 and an inductive argument, we find that $g(x,n)=x^n$ for all $x\ge0$ and all positive integers $n.$ Applying 5, 1, and another inductive argument, we find that $g(x,n)=x^n$ for all $x>0$ and all integers $n.$ Applying Archimedean property and 6, we have $\langle x,y\rangle\in E$ whenever $x>0,$ and that $\langle 0,y\rangle\in E$ whenever $y\ge 1.$ Applying 4 lets us conclude that $g(0,r)=0^r$ for all positive rational $r,$ whence another application of Archimedean property and 6 let us conclude that $\langle 0,y\rangle\in E$ whenever $y>0.$ Two applications of 4 let us conclude that $g(x,r)=x^r$ whenever $x>0$ and $r$ is rational--the first gets us to $\bigl|g(x,r)\bigr|=x^r,$ the second gets us the rest of the way. Applying 7, we find that $g(x,y)=x^y$ whenever $x>0,$ and that $g(0,y)=0^y$ for all $y>0.$ Another application of 7 shows that $\langle 0,0\rangle\notin E,$ whence an application of 6 shows that $\langle 0,y\rangle\notin E$ for $y<0,$ and a final application of 6 lets one show that $\langle x,y\rangle\notin E$ for $x<0.$

Instead of assuming convexity of $E$, we could assume that $E$ is connected, as is each set of the form $E\cap\bigl(\{x\}\times\Bbb R\bigr)$ for $x\in\Bbb R.$ However, connectedness of $E$ alone is insufficient. For example, $g(x,y)=x^y$ and $E=\bigl\{\langle x,y\rangle\in(\Bbb R\times\Bbb Q)\cup(\Bbb Q\times\Bbb R):(x>0)\vee(x=0\wedge y>0)\bigr\}$ satisfy 1 through 5, 7, and connectedness of $E,$ but not all $\langle x,y\rangle$ with $x>0$ lie in $E.$

Another observation is that weakening 3 to say "$x>0,$" we can conclude that $g(x,y)=x^y$ for all $\langle x,y\rangle\in E,$ and that $\langle x,y\rangle\in E$ whenever $x>0,$ and we needn't even apply condition 4. However, we no longer have uniqueness of $E.$

One final observation is that the functional equation of condition 4 is redundant, as it follows from condition 1, so we can safely remove it.

My Questions: Condition #3 feels artificial to me. Is there a way to replace it with a weaker one (as I mentioned that we can do with condition 4), possibly in conjunction with some other "natural" conditions like 1 and 2?