If $L$ is NOT Turing-recognizable, does that mean that super sets with $L$ are also not?
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.