Which is the fastest way to find a member of a subgroup with known size modulo prime $P$, with know factorization of $P-1$?->$x$ with $x^s \mod P = 1$ Assuming you know the factorization of used prime $P-1$
$P-1 = s \cdot f_2\cdot f_3...f_i$
Now you want to find a member of a subgroup $\mathbb{Z}_s$. 
This means any $x$ with
$x^s \equiv 1 \mod P $  

Naive way whould be selecting a random value $x$, compute $x^s$ and check if it is equal to $1$.

Another way I found online: 
transform to disc. log
This first first computes a prime root $g$ of $P$. With this you can rewrite the equation:
$x^s \equiv (g^k)^s \equiv  (g^s)^k \mod P$
And computes a $k$ with Shanks' baby-step giant-step algorithm.

Is that the best/fastest way to go? Does it help if $s$ is a prime?
Is there a faster way if you are allowed to change $s$ and $P$ as well?
e.g. instead finding $x$ look for a fitting P' instead. For this fix $x$ to any number of choice $x_{const}$ and search for a  $P'=s \cdot f +1$
$x_{const}^s \equiv 1 \mod P' $ 

(Trivial $x$ like $x = 1+n \cdot P$ do not count here)
Is that the best/fastest way to go?


Well, unless you give some criteria, whether it's the best is unanswerable.  However, it might not be the fastest; you could just note that $k = f_2 \cdot f_3 \cdot … \cdot f_i$ and skip the baby-step-giant-step algorithm entirely.

In addition, if all you want is a random element of the subgroup, you don't need to find a generator.  Instead, all you need to do is select a random value $r \in [1, P-1]$, and compute $x = r^k \bmod P$ (using the above definition of $k$); if $x \ne 1$ (true with probability $1 - 1/s$), that's what you're looking for.

BTW: why do you think you need $x$ to be a prime?