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Python 2, 68 bytes

One of the few cases where itertools is useful.

lambda b,a:s(b)in map(s,product(*a))
s=sorted
from itertools import*

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Python 2, 70 bytes

f=lambda s,l:l and any(f(s.replace(x,'',1),l[1:])for x in l[0])or''==s

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Tries every possible character from each list entry in turn, removing it from the target string s if present, and checking if we end up with the empty string. The s.replace(x,'',1) removes at most one copy of x in the string s.

An annoying issue is that in the base case where l==[], we get an error when we try to read l[0] even if the result doesn't matter, making it hard to cut the l and. It "almost" works to save a byte as below, but we get an error when l becomes empty because the l[0] is evaluated before the if l is tested and fails.

f=lambda s,l:s==''or any(f(s.replace(x,'',1),l[1:])for x in l[0]if l)

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69 bytes

f=lambda s,l:any(f(s.replace(x,'',1),l[1:]+[[]])for x in l[0])or''==s

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Pad l to the right with empty lists.

69 bytes

f=lambda s,h,*t:t and any(f(s.replace(x,'',1),*t)for x in h)or s in h

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Takes input as the target string, followed by a splatted list of lists.

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69 bytes

f=lambda s,l:any(f(s.replace(x,'',1),l[x in l[0]:])for x in s)or[]==l

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A "dual" method based on ovs' solution, iterating over characters in the target string and removing from the list.

Haskell, 54 bytes

import Data.List
(.permutations).any.flip elem.mapM id

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Without imports, 85 bytes

x#[]=[[x]]
x#(a:b)=(x:a:b):map(a:)(x#b)
(.foldr((=<<).(#))[[]]).any.flip elem.mapM id

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