Is a language whose Turing Machine doesn't halt for some positive cases but for others does not recursive?
There are two important misunderstandings in your question.


You talk about "the Turing machine" for a language but there isn't just one: in fact, if a language is recursive (or RE) then there are infinitely many Turing machines that decide (or accept) it.
If a Turing machine decides a language&nbsp;$L$ then, by definition, it accepts every input in&nbsp;$L$ and rejects every input not in&nbsp;$L$ and, therefore, it halts on every input. It doesn't make sense to talk about a Turing machine deciding a language but not halting on some input, or not accepting some string that's in the language.


With that in mind, a language&nbsp;$L$ is:


Recursively enumerable if there is a Turing machine that accepts every string in&nbsp;$L$ and does not accept any string not in&nbsp;$L$. It may loop or reject strings not in&nbsp;$L$.
Recursive if there is a Turing machine that accepts every string in&nbsp;$L$ and rejects every string not in&nbsp;$L$.
Recursively enumerable but not recursive if there is a Turing machine that accepts every string in&nbsp;$L$ and does not accept any string not in&nbsp;$L$, but no TM that accepts everything in&nbsp;$L$ and rejects everything not in.  The "but..." part can be rephrased as "every Turing machine that accepts every string in&nbsp;$L$ must either also accept some string not in&nbsp;$L$ or must loop on some string not in&nbsp;$L$."
Not recursively enumerable if every Turing machine either accepts at least one string not in&nbsp;$L$, or loops or rejects at least one string in&nbsp;$L$.


For your specific language, I'm going to assume that $\Sigma=\{a,b\}$. If not, then the notation gets a little uglier but nothing fundamental changes.

If $L$ is RE but not recursive, then
$$L' =\{aw\mid w\in L\} \cup \{bw\mid w\notin L\}$$
is not RE. suppose it is RE and is accepted by some Turing machine&nbsp;$M$. Then we can decide if $w\in L$ by simulating $M(aw)$ and $M(bw)$ in parallel until one of them accepts.  If $M$ accepts&nbsp;$aw$, then $w\in L$; if it accepts&nbsp;$bw$, then $w\notin L$, and one of these two things must happen. This contradicts the assumption that $L$&nbsp;is not recursive.