Suppose $\mathcal{M} \models T$ where $T$ is a set of $\mathcal{L}$-sentences. Now create an augmented language by adding all the constants in $\mathcal{M}$:
$$\mathcal{L}^* = \mathcal{L} \cup \{c : c \in \mathbb{M}\}$$
Then create a new theory $T^*$ using the language $\mathcal{L}^*$. Does $T^*$ contain any proposition that is materially different than $T$? I would think no because $T^*$ simply states the same propositions, but is now giving names to the elements, correct? I would think $\mathcal{M} \models T^*$. But is it possible now to find a different model $\mathcal{N} \models T^*$ but $\mathcal{N} \not\models T$?
I think your "$T^*$" is intended to be the elementary diagram of $\mathcal{M}$? http://en.wikipedia.org/wiki/Elementary_diagram
If this is correct, then "$\mathcal{M}\models T^*$" is almost true; the issue is that $\mathcal{M}$ is an $\mathcal{L}$-structure, not an $\mathcal{L}^*$-structure. So really what we want to say is that the obvious expansion $\mathcal{M}^*$ of $\mathcal{M}$ is a model of $T^*$.
On the other hand, and I think this might address your question, it is not the case that any $\mathcal{N}\models T$ can be expanded to a model of $T^*$, even if $T$ is complete! For example, let $T$ be the true theory of arithmetic, $\mathcal{M}$ be a nonstandard model of $T$, and $\mathcal{N}$ be the standard model of $T$. Then $\mathcal{L}^*$ contains - among other things - a symbol $c$ which corresponds to a non-standard element of $\mathcal{M}$, so $T^*$ contains the sentences "$c\not=0$," "$c\not=1$," . . . But clearly there is no possible assingment of $c$ to any element of $\mathcal{N}$ that would satisfy all of these statements at once.
(Note that this really shows that we should write "$\mathcal{L}_\mathcal{M}$," "$T_\mathcal{M}$," etc. instead of using just $*$, since it's important to indicate what model is being used to construct the expansion.
EDIT: I'm assuming your last sentence is a typo: since $T^*\supseteq T$, any model of $T^*$ is (the expansion of) a model of $T$, trivially.