The phrasing of your question is a little peculiar: one would usually impose the condition that $D$ is square-free. You also imposed the condition that the norm of $\varepsilon$ is $+1$, whereas the fundamental unit may have norm $-1$.

Under the assumption that $N(\varepsilon) = 1$, there is a trivial equality:

$$\varepsilon + \overline{\varepsilon} = a.$$

From the assumption that $\varepsilon > 1$, this yields the inequality

$$a - \frac{1}{a} > \varepsilon > a - \frac{1}{a-1}.$$

From this is follows that $|\varepsilon - \varepsilon'| > 1$ for distinct units of norm $+1$. Thus $M(1)$ is finite. For any $a > 2$, there exists a unique $\varepsilon$ with norm $+1$, given by solving the following equation for $b^2 D$:

$$a^2 - b^2 D = 4.$$

This specifies $\varepsilon$ uniquely. An estimate similar to the one given in quid's answer (in fact, it can be derived directly from that estimate) will show that the number of fundamental units which have norm $+1$ with $a \sim \varepsilon \le x$ is $x + O(x^{1/2})$, and thus that $M(T)$ is infinite for $T > 1$. The disparity between this answer and the one above is whether one allows units of norm $-1$ or not. (This estimate also shows that all but $O(x^{1/2})$ of the units $\varepsilon$ generated in this way for any $a \le x$ are fundamental units.)

If you wish to allow units of norm $-1$, then note that for every $a > 2$ there exists a unique unit $\varepsilon > 1$ of trace $a$ and norm $1$, and a unique unit $\epsilon > 1$ of trace $a$ and norm $-1$. We saw that above in the first case, and the second case is the same: simply solve the equation:

$$a^2 - b^2 D = -4.$$

In this case we will still have

$$\epsilon + \overline{\epsilon} = a,$$

and hence
$0 = (\varepsilon + \overline{\varepsilon}) - (\epsilon + \overline{\epsilon})$, from which we get

$$|\varepsilon - \epsilon| = |\overline{\varepsilon} - \overline{\epsilon}| \sim \frac{1}{2a}.$$

From the reference in quid's answer, we know that for $x + O(x^{1/2})$ of the $a \le x$, either (and hence, because $O(x^{1/2}) + O(x^{1/2}) = O(x^{1/2})$, both) $\varepsilon$ and $\epsilon$ are fundamental units. Hence, allowing units to be of norm $-1$ and modifying your definition of $M(T)$ appropriately, one has $M(\delta) = \infty$ for any $\delta > 0$.