Prove that regular expression is unambiguous
This answer has 3 parts:


I first answer the question as asked,
sketching the beginning of a proof concerning the example given. A
full proof is too long and tedious to be given here.
Then I give a general procedure to solve the problem and decide
whether a regular expression is deterministic/unambiguous.
Finally I discuss some difficulties and problems with the statement
of the question.


A direct answer to the question

What you may first prove, though it should be obvious, is that for any
regular expression $R$, there is a valid mapping of any word $w\in L(R)$
to $R$.

Then you consider the regular expression $R = 0^*(11^*000^*)^*11^*01(1+0)^*$ to be analyzed, and you
procede to decompose it structurally, showing  that at every step,
there is only one way to decompose a string $w\in 

For example, you show that there is only one way to decompose a string
$w\in L(R)$ as $w=uv$ with $u\in L(0^*)$ and $v\in
L((11^*000^*)^*11^*01(1+0)^*)$. The reason is that $u$ contains only
$0$, and $v$ begings necessarily with a $1$. So the string $w$ must be
cut just before the first one. Then there is only one way to get a
valid mapping for the whole expression, from valid mapping of the two
subexpressions. If the subexpressions are themselves proved
detrministic, i.e. with only one valid mapping fo any string in their
language, then the whole expression must be deterministic.

Hence you now have to prove only that, $0^*$ and
$(11^*000^*)^*11^*01(1+0)^*$ are deterministic, since their
concatenation has a unique decomposition.

So you go on decomposing the expression into subexpression, such that
you have a unique way of matching them with a string of the language.

The next step is to show there is only one way to decompose a string
$w$ in $L((11^*000^*)^*11^*01(1+0)^*)$, as $w=u01v$, with $u\in
L((11^*000^*)^*11^*)$ and $v\in L((1+0)^*)$

And so on ... For each subexpression, you must perform the analysis
according to the type of the subexpression, following the definition
given for each of the four types in the definition of valid mapping.

In the case of $R=S+T$, where  $S$ and $T$ are subexpressions of $R$, you only have to prove that $L(S)\cap L(T)=\emptyset$.

This gives you the general idea of a proof on a given example, such as the one above.

However, such a proof is very tedious and prone to errors.

So you would be better off designing an algorithm that will decide mechanically whether a regular expression is ambiguous or not. The problem is decidable, and not so hard. I follows the same lines I sketched: recursive decomposition of the regular expression, which things to be checked depending on the operator (concatenation, union, Kleene star).

A procedure to decide ambiguousness of a regular expression.

To decide ambiguousness of a regular expression, one must
recursively decompose the expression into 1 or 2 subexpressions
dominated by an operator, down to the leaf level.

Given a regular expression $R$,


If $R=w\in \Sigma+$, then $R$ is unambiguous;
If $R=R_1+R_2$, then $R$ is unambiguous iff both $R_1$ and $R_2$
are unambiguous, and $R_1\cap R_2=\emptyset$. If the intersection is
not empty, any valid mapping from string to $R_1$ could instead map
validly to $R_2$; hence there would be two valid mappings. Also, if
there were two distinct valid mappings to one daughter, say $R_1$,
then by application of rule 2 of the definition, there would be two
valid mappings to $R$.
If $R=R_1 R_2$, then $R$ is unambiguous iff both $R_1$ and $R_2$
are unambiguous, and there is only one way to decompose any string
$w\in L(R)$ into 2 substrings as $w=w_1 w_2$ such that $w_1\in
 L(R_1)$ and $w_2\in L(R_2)$. $R_1$ and $R_2$ must be both unambiguous
because, as before, if one were ambiguous, say $R_1$ for a string
$w_1$, if would be possible to build two valid mappings for $R$,
which would thus be ambiguous.

But how does one check that string $w\in L(R)$ can be decomposed in
only one way into two substrings 
$w=w_1 w_2$ with $w_i\in L(R_i)\text{ for }i=1,2$? This can be
checked by attempting to simulate 2 simultaneous generations of a
string in $L(R)$, such that 2 distinct prefix of the substring will
be associated to $R_1$.

Let $A_1$ and $A_2$ be two FA recognizing $L(R_1)$ and $L(R_2)$,
having (to simplify) one start state and one accepting state.
A FA for $L(R)$ can be obtained by linking then with an empty
transition from the accept state of $A_1$ to the start state of
$A_2$.

Then one can build a new FA $A^2$ by doint a cross product of the
state set of $A$ with itself to simulate 2 computations at the same
time. Both computations must be synchronized so that they scan the
input together, though not necessarily following the same
computational paths for each simulated automaton. Both computations
must at some point cross the linking transition. This is memorized in
the final state so that the double computation ends in acceptance iff
the both reach the final state of $A_2$, but after crossing the
linking transition at different points in the input.

This automaton $A^2$ recognizes the language $\{w\in L(R)\mid
 w=uv=u'v' \wedge u\neq u' \wedge u,u'\in L(R_1) \wedge v,v'\in L(R_2)$

Hence, strings in $L(R)$ can be decomposed unambiguously in two
strings belonging respectively to $L(R_1)$ and $L(R_2)$ iff
$L(A^2)=\emptyset$.
If $R=S*$, then $R$ is unambiguous iff $S$ is unambiguous, and
there is only one way to decompose any string $w\in L(R)$ into a
concatenaton of substrings in $L(S)$.

The technique is similar to the one used for concatenation. You take
a FA $A_S$ recognizing $L(S)$, with a single accept state, and you
connect the accept state to the start state with an empty "looping" transition,
to get a FA $A$ for $L(R)$. Then, you buils a FA $A^2$ that simulates
two computations of $A$ and accepts only when they both accept, but
with the constraint that they must have crossed the looping
transition at least once while being at different points of the input
(or possibly a different number of times).

Again, strings in $L(R)$ can be decomposed unambiguously in a
sequence of strings in $L(S)$ iff $L(A^2)=\emptyset$.


Caveats

Some care must be taken with the empty word $\epsilon$. It is not
clear that the problem is always well defined in this respect.
The empty word $\epsilon$ may be used in regular expressions but will have no
apparent counterpart in strings of the language. They can be ignored,
But then, what of a Kleene star $R=S^*$ when  $\epsilon\in L(S)$.
Should it be considered always ambiguous.

Another point is that the concept of valid mapping is really a mapping
of occurrences of words to occurrences of regular expressions, i.e., to
the syntactic instances, rather than to regular expressions as
abstract concepts.

For example, consider the string $w=011001$ and the regular expression
$R=(01)^*100(111)^*1$. There is a valid mapping $f$ of $w$ to
$R$. However it associates the first instance of $01$ with $(01)^*$,
and the second instance of $01$ with $0(111)^*1$.

What is worse, even considering occurences, $f$ is not a mapping since
it associates the first instance of $01$ with both $(01)^*$ and the
regular expression $01$ to which the Kleene star is applied. Hence
$f$, as it is defined, cannot be considered a mapping.

Actually, it seems that $f$ should be defined as a mapping of $R$ and each
subexpression occurrence in $R$ to
substrings of $w$.

But up to that rephrasing of the problem, everything holds.