One possible definition of recursively enumerable (most likely the one
used by your instructor) is that it is the domain of a partial
function. In Turing Machine terms, that means the following
definition:
A language $L$ is recursively enumerable iff there is a turing
machine that halts on input $x$ exactly whenever $x\in L$.
Note that there is no longer any notion of accepting an rejecting
state.
Now, assuming that you have a recursive language $L$, you know there
is a TM $M$ that always halt on any input $x$ (on the alphabet of $L$), being
in an accepting state iff $x\in L$.
To prove that L is RE, you can build a machine $M'$, that will halt
exactly on the words $x\in L$. This implies the following step, if you
want to be fully formal:
define the TM $M'$;
prove that $M'$ halts on all words $x\in L$;
prove that $M'$ halts only on words $x$ that are in $L$, i.e. does not halt on words $x\notin L$.
Actually, in most proofs of this type, the difficulty is to define
(construct) the machine $M'$, while the two other steps are often long and
tedious, but not difficult. So it is common to consider that giving
the definition is enough of a proof (as long as the definition is correct).
However, that may be unwise, because sometimes the other two steps
will uncover errors in the definition provided in the first step. So you are
quite correct in asking for a full proof. The definition of $M'$ is
not enough for a formal proof, even though it is pretty trivial in the
example of your question.
Actually, the construction you give, possibly badly recorded, does
contain two error:
Given the TM $M$ that can decide whether a string $x$ belongs to $L$,
you define the TM $M'$ as follows:
convert all non-accepting halting states into states that loop for
ever;
convert all accepting halting states into simple halting states (not really
important, but acceptance will be irrelevant, only halting will
matter);
Everything remains otherwise the same: same states, same transitions,
and you have only added transitions to make the non-accepting halting
states go into a loop.
Then, you can prove by induction on the length of a computation
(number of transitions), that if $M$ reaches an instantaneous
description $d$ on input $x$ in $n$ steps, then $M'$ reaches the same
$d$ in input $x$ also in $n$ steps. Whenever $x\in L$, there is a
computation of $M$ such that $M$ reaches an instantaneous description
with some halting accepting state $q$. Hence, the same computation for
$M'$ must reach the same instantaneous description with state $q$,
which by construction of $M'$ is still a halting state. Therefore $M'$
is in a halting state.
When $x\notin L$, there is a computation of $M$ such that $M$ reaches
an instantaneous description with some halting non-accepting state
$q$. Hence, the same computation for
$M'$ must reach the same instantaneous description with state $q$,
which by construction of $M'$ is a looping state, so that $M'$ will
not terminate.
So the TM $M'$ is a machine that does recognize $L$ as an RE set,
according to the definition above.
QED