Without knowing about logarithms, the question need not even begin to make sense, since it is unclear just what $a^x$ might mean, even for positive $a$, when $x$ is not restricted.
You say in the comments that you would define the function $f(x)=a^x$ simply as a function with the property that $$\begin{align*} f(1) &= a\\ f(x+y) &= f(x)f(y). \end{align*}$$
However, these properties do not suffice to conclude that $f$ is one-to-one (which is required in order to conclude that $f(x)=f(y)$ implies $x=y$.
In particular, if we assume the Axiom of Choice, then there are functions that satisfy both $f(1)=a$ and $f(x+y)=f(x)f(y)$, but that are not one to one.
"Explicitly" (modulo the Axiom of Choice), let $\beta$ be a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ such that $1\in\beta$. Then any function $g\colon \beta\to\mathbb{R}$ can be extended to an additive function $g\colon\mathbb{R}\to\mathbb{R}$; that is, a function defined on all of $\mathbb{R}$, whose values at $\beta$ are as specified, and such that for all $x,y\in\mathbb{R}$, we have $g(x+y)=g(x)+g(y)$.
Now, define $g\colon \beta\to\mathbb{R}$ by letting $g(1) = 1$ and $g(r)=0$ for all $r\in\beta$, $r\neq 1$. Then define $f\colon\mathbb{R}\to\mathbb{R}$ by $f(x) = a^{g(x)}$.
Then $f(1) = a^{g(1)} = a= a$; and $f(x+y) = a^{g(x+y)} = a{g(x)+g(y)} = a^{g(x)}a^{g(y)} = f(x)f(y)$. So this function $f$ satisfies the two given equations.
However, $\beta$ is uncountable, so pick $r\neq 1$ that is in $\beta$. Then $f(r) = a^{g(r)} = a^0 = 1$, and $f(0) = 1$ (since $g(0)=0$ must hold for $g$ to be additive). However, $r\neq 0$, since $0$ cannot be an element of $\beta$.
Thus, the two conditions $f(1)=a$ and $f(x+y)=f(x)f(y)$ do not suffice to show that $f$ is one-to-one.
Which means you need to specify a lot of other stuff; specifically, one need to know exactly what properties you are giving the function $f$.
It is difficult to define either the exponential or the logarithm at the basic level of calculus-before-the-fundamental-theorem.
One can define the exponential function by first defining the functions $a\longmapsto a^n$ with $n$ a positive integer, inductively. Then for $n$ a negative integer using reciprocals. Then for $n$ the reciprocal of a positive integer using inverse functions. Then for $n$ a rational using $a^{p/q} = (a^p)^{1/q}$. Then prove that if $\frac{p}{q}=\frac{r}{s}$ with $p,q,r,s$ integers, $r,s\gt 0$, then we get $a^{p/q}=a^{r/s}$. Then define $a^x$ for arbitrary $x$ by letting $(q_n)$ be a sequence of rationals such that $q_n\to x$ as $n\to\infty$, and showing that the sequence $a^{q_n}$ is Cauchy and converges to a number we call $a^x$. Then showing that if $(q_n)$ and $(r_n)$ are two sequences of rationals that both converge to $x$, then $\lim_{n\to\infty} a^{q_n} = \lim_{n\to\infty} a^{r_n}$. And once all of this has been done, then one can show that the function is stricitly monotone wheN $a\gt 0$, $a\neq 1$, to deduce what you want (and hence that it has an inverse and logarithms exist).
Obviously, this requires a lot of work.
Or one can use integrals and define the natural logarithm by $$\ln(x) = \int_1^x \frac{1}{t}\,dt$$ for $x\gt 0$. Using the Fundamental Theorem of Calculus one can show that this function is continuous and differentiable; using the properties of the integral that it is strictly increasing, and so has an inverse. Call the inverse the exponential function $\exp(x)$; and then define $a^x = \exp(x\ln(a))$. And then prove that this function is strictly monotone when $a\gt 0$, $a\neq 1$.
This requires enough Calculus to prove the Fundamental Theorem of Calculus first. Again, a lot of work.
