Logical extension of exponential function?

Question

I’ve been thinking a lot about logical extensions of the exponential function, i.e. one for multiplication and addition (and maybe even one for tetration?)

We know (almost by definition) that: $$ \sum_{i=0}^{\infty}\frac{\left(\log_e\left(a\right)x\right)^{i}}{i!}=a^x $$

And I discovered: $$ \sum_{i=0}^{\infty}\frac{\left(\frac{ax}{e}\right)i}{i!}=ax $$

However: $$ \sum_{i=0}^{\infty}\frac{\left(a+x-e\right)+i}{i!}\neq a+x $$

Can anyone think of another logical extension from the exponential function? Whether for multiplication or for addition?

Thanks!

Post script: I also discovered this: $$ \frac{\sum_{i=0}^{\infty}\frac{\left(a+x-e\right)+i}{i!}}{e}+e-1=a+x $$ though this is certainly less elegant and doesn’t appear to be a logical extension.

Of possible interest is my 30 January 2006 sci.math post "Is e^x zero in some extended number system?" and its same-day follow-up. However, I know almost nothing about the kind of generalization discussed there. The paper I excerpt from is Paul Dienes, The exponential function in linear algebras, Quarterly Journal of Mathematics (Oxford) 1 (1930), 300-309 — Dave L. Renfro, Feb 12 '21 at 13:45
@DaveL.Renfro Definitely interesting. though I can’t claim I understood much of it. I’m also not entirely sure how it’s related... — Rowan Ackerman, Feb 12 '21 at 13:54
There's also the notion of exponentiation of an operator (an operator being a function in which "ordinary functions" are the inputs and outputs, such as the differentiation operator), and things like $e^{\frac{d}{dx}}$ actually play an important role in quantum mechanics. See also my answer to Is there a such thing as an operator of operators in mathematics? — Dave L. Renfro, Feb 12 '21 at 14:01
So you have $f_a(x)\in{a^x,a\cdot x,a+x}$ and you write $$f_e(f_e^{-1}(f_a(x))=f_a(x)$$ while using Taylor expansion for $e$, then you rewrite RHS to put (almost) everything under the sum? And your question is, "Can anyone think of another way to rewrite RHS?" ? — Vepir, Feb 12 '21 at 15:00
@Vepir No, my question is: Why, logically, doesn’t the third equation in my post work, and what would logically make it work. I will stress, I don’t mean Mathematically, as I have a Mathematical solution, I mean logically. — Rowan Ackerman, Feb 12 '21 at 17:31
The fact that multiplication is distributive over addition is what makes the second equation work and is what makes the third equation fail. — Vepir, Feb 12 '21 at 18:12

Logical extension of exponential function?

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