If $L$ is NOT Turing-recognizable, does that mean that super sets with $L$ are also not? Suppose there is a language $L$ that is not Turing-recognizable. Now suppose, for every word in $L$, you would concatenate a "0" as prefix, resulting in $L' = \{0w | w \in L\}$. Does this mean that $L'$ is also not Turing-recognizable?

I would argue that $L'$ is also not Turing-recognizable:

No Turing machine recognizes $L$. So there exists no TM that recognizes all the words in $L$. That means that all the words that can not be recognized by a TM that are in L, also can not be recognized even with a "0" added before them. Because after the TM's head reads the "0", we assume that the head will read what comes after, which is a word that is not Turing-recognizable. One could argue that the head doesn't need to read the word and just go to "accept" after reading the 0, but then I can also build a TM that would just go to "accept" straight away before reading this word at all, meaning that the word was not Turing-recognizable which contradicts the initial statement.

In other words, after reading the "0", the TM will have to come across the not Turing-recognizable word, which it can't recognize. Therefore $L'$ is also not Turing-recognizable.

To generalize, I think if a not Turing-recognizable language is a subset of another language, then that language is also not Turing-recognizable.

However, I feel my explanation is shaky and I'm not convinced I am correct. Can anyone verify my claim, and either prove me wrong or give tips for a better proof?
If $L'$ were Turing-recognizable then there would be a TM $M$ which would enumerate every string in $L'$. In particular, we could construct a TM that would print on the tape every string in $L'$. Then we could easily modify this machine to enumerate the language $L$. Just do not print the first symbol on the tape (skip the initial $0$). Thus we construct a machine which enumerates $L$. But $L$ is not Turing-recognizable (not r.e.) by assumption. A contradiction. 

$L'$ in your example may be expressed as a concatenation of languages $\{0\}$ and $L$, that is, $L' = \{0\}L$. However, we cannot generalize this result to arbitrary languages. For example, $\{0,1\}^*L = \{0,1\}^*$ is r.e. even if $L$ is not r.e. language on $0$ and $1$ containing empty string.

As for your claim "a not Turing-recognizable language is a subset of another language, then that language is also not Turing-recognizable", it is false too, since a not Turing-recognizable language $L$ on $0$ and $1$ is a subset of $\{0,1\}^*$ which is definitely r.e., even decidable.