How to define the exponential function without calculus?

You won't be able to do derive

$exp(1)=e$

from your definition, since your definition works for the exponential function for any base $b$:

$b^0=1$

and

$b^{x+z}=b^x\cdot b^z$

are true for any $b$!

As Bram28 alludes, your definition is invariant under scalar multiplication. That is, if $u=cx$ for some $c$, then both of your conditions work just as well for $u$ as for $x$. For instance, $f(2(x+y)) = f(x+y)f(x+y) = f(x)f(y)f(x)f(y)$. And $f(2x)f(2y) = f(x)f(x)f(y)f(y)$. So if $f(x+y) = f(x)f(y)$, then $f(2(x+y)) = f(2x)f(2y)$. There is therefore no way to distinguish between non-zero numbers. If you have an argument for why $f(1) = e$, I can just multiply everything by two and everything will work the same as before, and I'll end up for an argument for why $f(2) = e$.

You have to take $f(1)$ as a constant, and then find $f(x)$ in terms of that constant. Then $f(n+1) = f(n)f(1)$, so by induction $f(n) = f(1)^n$ for natural number $n$, and a similar argument gets negative integers. You can then argue that $f(1) = f(.5+.5)=f(.5)f(.5)$, so $f(.5) =\sqrt{f(1)}$ (assuming $f(1)$ is positive). It's then not too difficult to get $f(x)$ defined for any rational number. For irrational numbers, though, you have to assume $f(x)$ is continuous.

$$\begin{align}\exp(x)&=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\&=\sum_{i=0}^{\infty}\frac{x^i}{i!} \end{align}$$