I am not an expert, but it seems to me that without requiring more from the family $L\to S$ this will not hold. (Well, you didn't really say what *family* means so requiring "more" is an understatement).

Here is an example to test your ideas and conditions on:
Suppose $M$ is really nice, but at least *normal* and suppose there exists a morphism $\alpha:M\to S$ which is one-to-one on closed points but not an isomorphism. Say $M=\mathbb A^1$, $S$ is a cuspidal cubic and $\alpha$ is the normalization. Now consider the *family* on $S$ obtained by composing $\alpha$ with the morphism $G\to M$. So you have essentially the same family, at least the same fibers (kinda) just a little cusp-ed at some points of $S$. 

Now if this is an admissible family in your situation (I guess it may not be as the fiber over the cusp will be a multiple fiber and you probably disallow that, but who knows), then you have a problem: If this family is to be pulled-back from $M$, then the desired morphism should be an inverse to $\alpha$ (at least point-wise), but we know that there is no such map as a non-normal point cannot dominate a normal one with degree one.

On the other hand, what you can certainly do in your original situation is to pull-back the family $L$ to the normalization of $S$ and obtain your map to $M$ that exhibits it as a pull-back. Then you can try to analyze the situation and see if this morphism from the normalization of $S$ would factor through $S$. The main question is whether a crooked family as above is possible.