As Peikert has commented (and my first answer was not dealing with this issue carefully), the LWE problem asks you to distinguish between
$(\vec a, \langle \vec a, \vec s \rangle + e \mod q) \in \mathbb{Z}_q^{n+1}$
and $\vec u \in \mathbb{Z}_q^{n+1}$ (with each entry of $\vec u$ sampled uniformly from $\mathbb{Z}_q$) given many samples.
This is why the LWE problem is usually defined with a parameter $m$, which represents the number of samples
$(\vec a_i, \langle \vec a_i, \vec s \rangle + e_i \mod q)$
that you can get. Notice that all the samples use the same fixed secret $\vec s$.
The problem says nothing about how you are going to distinguish those two distributions. Given many samples, if you run an statistical test or some clever algorithm using lattices, it does not matter, as long as you can tell to which distribution those samples belong.
And about $b_i := \langle \vec a_i, \vec s \rangle + e_i \mod q$ being distributed uniformly on $\mathbb{Z}_q$, indeed, if you have a single sample, then that is the case, but again, if you have several samples, say, $m$,
then, the distribution of $(a_i, b_i)_{i=1}^m$ can possibly be distinguished from the uniform over $\mathbb{Z}_q^{n+1}$.
In particular, if you can recover $\vec s$, then you can simply compute $b_i - \langle \vec a_i, \vec s \rangle \mod q$ to get $e_i$, which are all small, thus, not uniformly distributed.
A final note, in your question you say "From what i understand a and e are both randomly sampled from uniform distribution", but actually, only $\vec a$ is uniformly chosen from $\mathbb{Z}_q$, the noise term $e$ usually follows a distribution that is likely to sample values much smaller than $q$ (like a Gaussian distribution with parameter approx. $\sqrt n$).
