The minimal $s$ is $3$.

It is attained by several elliptic K3's,
including $y^2 = x^3 + (t^2-t)^4$ which has IV* fibers at
$t = 0, 1, \infty$ and no other singular fibers. 
The comment by **Ariyan Javanpeykar** gives one argument that 
$s$ can be no smaller. (*See postscript.*
This uses characteristic zero; in small positive characteristic
$s$ can be as small as $1$, e.g. in characteristic 2 the elliptic K3 surface
$y^2 + y = x^3 + t^9$ has only one reducible fiber, at $t = \infty$.)

P.S. There are other ways to prove $s>2$; for example it follows from
Szpiro's inequality, which has an elementary proof via the 
Mason-Stothers theorem (polynomial ABC).  See MO 190530,
<https://mathoverflow.net/questions/190530>, for this and
some related ideas.  (That question is related because
${\bf G}_m({\bf C}) = {\bf CP}^1 - \{0, \infty\}$ and
if an elliptic surface $\pi: X \to {\bf P}^1$ has $s \leq 2$
then one can choose the coordinate on ${\bf P}^1$ so that
each bad fiber maps to $0$ or $\infty$.)