Prove $\ln(a \cdot b)=\ln(a)+\ln(b)$, $a,b>0 $ and $\lim_{a \to 0} \frac{\ln(1+a)}{a} =1$.But we shall not even know that the exponential function.

It's really good that you are noting this must be proven and not taken for granted!

Integration by substitution is valid. So your proof is valid.

Note if you use Reimann definition of integral

$\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^nf(x_i)\Delta x$ where we divide the interval $[a,b]$ into $n$ equal parts, $\Delta x = \frac {b-a}n$ and $x_i \in $ the $i$th of the smaller intervals (for practical purposes we may assume, assuming the function is integrible at all, the endpoint).

In other words: $\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f(a + i\frac {b-a}n)\cdot \frac {b-a}n$.

So

$\int_{a}^{b} \frac 1x dx = \lim_{n\to\infty} \sum_{i=1}^n\frac 1{x_i}\Delta_1 x=$ (where $\Delta_1 x = \frac {b-a}n$ and $x_i = a + i\Delta_1 x$. Now let $k$ be any non-zero constant....)

$\lim_{n\to\infty} \sum_{i=1}^n\frac 1{kx_i}k\Delta_1 x= $(where $k\Delta_1 x = \frac {kb- ka}n$ and $kx_i = ka + i(k\Delta_1 x)$)

$\lim_{n\to\infty} \sum_{i=1}^n\frac 1{y_i}\Delta_2 y= $(where $\Delta_2 y = k\Delta_1 x = \frac {kb-ka}n$ and $y_i = ka + i\Delta_2 y$)

$\int_{ka}^{kb} \frac 1y dy= \int_{ka}^{kb} \frac 1x dx$

So for any non-zero constant, $k$ we have $\int_a^b \frac 1x dx = \int_{ka}^{kb} \frac 1x dx$.

(we can give a similar argument for the other definitions of integral)

And with that....

Here's a different proof of the formula $\ln(ab)= \ln(a) + \ln(b)$.

Suppose that $a$ is a positive constant.

And suppose $f$ is the function defined over the positive reals by:

$f(x) = \ln(ax) - \ln(x) - \ln(a)$.

$f$ is differentiable over the positive reals. By the chain rule,

$f'(x) = a \frac{1}{ax} - \frac1{x} = 0$

So $f$ is constant.

As such, for all positive reals $x$, $f(x) = f(1) = \ln(a) - \ln(a) -\ln(1) = 0$.

Which concludes our proof.