You are correct: the regexp you quote does not match the string $10011$ and that string has the property that every pair of adjacent $0$s occurs before any pair of adjacent $1$s. But it does match the string $001100$, which fails to have the property!
Your proposed solution matches $11001$ so is also incorrect (the first bracket matches $1$, the second matches $10$, the third matches $01$ and the fourth matches $\epsilon$.
A regexp that does match the required language is
$$0^*(100^*)^*(11^*0)^*1^*\,.$$
It works as follows.
$0^*(100^*)^*$ matches the empty string and any non-empty string that doesn't contain $11$ and that ends with $0$. The string can't contain $11$ because there must be at least one $0$ between every two $1$s.
$(11^*0)^*1^*$ matches every string that begins with $1$ and does not contain $00$, since there must be at least one $1$ between any two $0$s.
Now put the two halves together. If a string has the property that all occurrences of $00$ are before the first $11$, then the section of the string before the first $11$ must either be empty or end in $0$, so it matches the first part of the regexp. The rest of the string contains no occurrences of $00$ so it matches the second part of the regexp.
Note that there are shorter regexps that do the same thing: Yuval Filmus gives $(0+10)^*(1+10)^*$ in a comment to the question. That's a tidier version of the same idea.
And, yes, $(0^*1^*)^* \equiv (0+1)^*$ matches every possible binary string.
