Spectral triples which are not $\theta$-summable

So, under this generality, it seems to me that we can choose $A$ in a clever way: if we make $A$ smaller, then it gets "easier" to satisfy the condition. So why not take $A$ to be the scalar multiples of the identity-- then $D$ commutes with all of $A$ on the nose.

Now let $H=\ell^2$ and let $D$ be multiplication by a sequence of real numbers $(d_n)$. If $z\in\mathbb C$ not an accumulation point of the $(d_n)$ then $(zI-D)^{-1}$ exists and is the multiplication operator by the sequence $(z-d_n)^{-1}$. If $d_n\rightarrow\infty$ then $(z-d_n)^{-1}\rightarrow 0$ and so $(zI-D)^{-1}$ is compact and so $D$ has compact resolvant.

Similarly, $e^{-tD^2}$ is the multiplication operator by the sequence $(\exp(-td_n^2))$. This will be trace class if and only if $$ \sum_n \exp(-td_n^2) < \infty $$ So you just need to let $(d_n)$ grow very slowly. For example, set $$ d_n = \big( \log(1/e_n) \big)^{1/2} \implies \exp(-td_n^2) = e_n^{t} $$ where we now just need that $e_n\rightarrow 0$. Let $e_n = 1/2$ for the first $N_2$ terms, then $e_n=1/3$ for next $N_3$ terms, and so on. Then $$ \sum_n \exp(-td_n^2) = \sum_{k\geq 2} \frac{N_k}{k^t}. $$ Pick $N_k \geq k^k$ so that for any $t>0$ if $K>t$ then $$ \sum_n \exp(-td_n^2) \geq \sum_{k\geq K} \frac{N_k}{k^t} \geq \sum_{k\geq K} 1 = \infty. $$

In this example, you could also take $A=c_0$ for a less trivial algebra.

One can produce an operator in a less artificial way. Consider the Sturm-Liouville operator $Du=-u''+V(x)u$, where $V(x)$ grows at infinity as $\log\log|x|$. The eigenvalues grow very slowly, also as $\log\log$, and therefore the exponent of the operator is not trace class (the details on the asymptotics of eigenvalues can be found, eg, in the book by S.Levendorskii, 'Asymptotic distribution of eigenvalues of differential operators. Kluwer Academic Publishers Group, Dordrecht, 1990.)