Perhaps a simpler, more tractable problem is, does there exist any compact connected abelian group which is not a product of $\mathbf{a}$-adic solenoids $\Sigma_\mathbf{a}$ for various sequence $\mathbf{a} = (a_1, a_2, \cdots)$ of integers greater than $1$ (cf. 10.12 and 10.13 ibid) and the circle group $\mathbb{T}$

Following YCor's comments, the Pontryagin dual question is whether there exists a torsion-free abelian group which is not a direct sum of localizations of $\mathbb{Z}$. And the answer is yes: for example, you can consider the $p$-adic integers $\mathbb{Z}_p$ as a *discrete* group. Cf. the accepted answer at this MO question about simply presented abelian groups; YCor's answer is also educational and suggests simpler examples such as suitable subgroups of $\mathbb{Z}[1/p]^2$.

**Edit:** Let me see if I can flesh out YCor's claim about subgroups of $\mathbb{Z}[1/p]^2$. The interesting ones are the ones of rank $2$, and for simplicity I'll restrict my attention to subgroups containing $\mathbb{Z}^2$ (I thought I had an argument that every subgroup of rank $2$ is isomorphic to such a subgroup but now I'm not so sure). These correspond to subgroups of the quotient $\mathbb{Z}[1/p]^2/\mathbb{Z}^2 \cong \mu_{p^{\infty}}^2$, the product of two copies of the Prüfer $p$-group.

We can describe such subgroups using Goursat's lemma, which says in this case that if $H \subseteq \mu_{p^{\infty}}^2$ is a subgroup such that the two projections $\pi_1, \pi_2 : H \to \mu_{p^{\infty}}$ are surjective (this is the interesting case), then there exist subgroups $N_1, N_2 \subseteq \mu_{p^{\infty}}$ and an isomorphism $\varphi : \mu_{p^{\infty}}/N_1 \cong \mu_{p^{\infty}}/N_2$ such that

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : \varphi(q_1) \equiv q_2 \bmod N_2 \}.$$

The proper subgroups of $\mu_{p^{\infty}}$ take the form $\mu_{p^n}$ for $n \in \mathbb{Z}_{\ge 0}$ (if we consider all of $\mu_{p^{\infty}}$ then $H$ is all of $\mu_{p^{\infty}}^2$). The quotient $\mu_{p^{\infty}}/\mu_{p^n}$ is isomorphic to $\mu_{p^{\infty}}$ again, and its automorphism group is the group of $p$-adic units $\mathbb{Z}_p^{\times}$, of which there are uncountably many.

So there are uncountably many choices for $H$, and hence uncountably many subgroups of $\mathbb{Z}[1/p]^2$ containing $\mathbb{Z}^2$. Of these, the subgroups isomorphic to $\mathbb{Z}^2, \mathbb{Z} \times \mathbb{Z}[1/p]$, or $\mathbb{Z}[1/p]^2$ are determined by the image of the copy of $\mathbb{Z}^2$ in each of them, and there are countably many choices for this image. So as promised, at most countably many subgroups can be isomorphic to a direct sum of localizations of $\mathbb{Z}$. Unfortunately I don't quite see how to explicitly exhibit a particular choice of $H$ which is not such a direct sum.

**Edit #2:** Let's take $N_1 = N_2 = 0$ to be trivial above, and $\varphi : \mu_{p^{\infty}} \cong \mu_{p^{\infty}}$ to be multiplication by a $p$-adic unit $u \in \mathbb{Z}_p^{\times}$. Then

$$H = \{ (q_1, q_2) \in \mu_{p^{\infty}}^2 : u q_1 = q_2 \}$$

is a subgroup of $\mu_{p^{\infty}}^2$ isomorphic to $\mu_{p^{\infty}}$, and it lifts to a subgroup

$$\widetilde{H} = \{ (q_1, q_2) \in \mathbb{Z}[1/p]^2 : u (q_1 \bmod 1) \equiv q_2 \bmod 1 \}.$$

Thinking of $H$ as a "line with slope $u$" suggests that $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ (in fact to $\mathbb{Z} \times \mathbb{Z}[1/p]$) iff $u$ is rational. Indeed, if $u = \frac{a}{b}$ is rational, so that $a, b$ are integers relatively prime to $p$ and each other, then the condition that $u (q_1 \bmod 1) \equiv q_2 \bmod 1$ is equivalent to the condition that $aq_1 - bq_2 \in \mathbb{Z}$, and consequently the map

$$\widetilde{H} \ni (q_1, q_2) \mapsto (q_1, aq_1 - bq_2) \in \mathbb{Z}[1/p] \times \mathbb{Z}$$

is an isomorphism.

Now suppose that $u$ is irrational. First let's show that $\widetilde{H}$ has no subgroup isomorphic to $\mathbb{Z}[1/p]$: equivalently, no nonzero element is $p$-divisible. If $(q_1, q_2) \in \widetilde{H}$ is a nonzero element, for it to be $p$-divisible would require that

$$u \left( \frac{q_1}{p^k} \bmod 1 \right) \equiv \frac{q_2}{p^k} \bmod 1$$

for all $k$, or equivalently that

$$u(q_1) - q_2 \in p^k \mathbb{Z}$$

for all $k$. Taking $k \to \infty$ gives $u = \frac{q_2}{q_1}$, but this contradicts $u$ irrational.

Hence in this case, if $\widetilde{H}$ is isomorphic to a direct sum of localizations of $\mathbb{Z}$ then it can only be isomorphic to $\mathbb{Z}^2$. But the projection to either coordinate gives a surjection onto $\mathbb{Z}[1/p]$, which $\mathbb{Z}^2$ does not possess.

Some additional comments. Projection onto the first coordinate shows that $\widetilde{H}$ is an extension

$$0 \to \mathbb{Z} \to \widetilde{H} \to \mathbb{Z}[1/p] \to 0.$$

The argument above shows that this sequence splits iff $u$ is rational. In general, this extension is classified by a class in $\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z})$, which can be computed from $\text{Ext}^1(\mu_{p^{\infty}}, \mathbb{Z}) \cong \mathbb{Z}_p$ and the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z}[1/p] \to \mu_{p^{\infty}} \to 0$ to give

$$\text{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \cong \mathbb{Z}_p/\mathbb{Z}.$$

The point of this computation is to show that for general homological reasons there are interesting nontrivial extensions of torsion-free abelian groups by torsion-free abelian groups; taking Pontryagin duals, there are interesting nontrivial extensions of compact connected abelian groups by compact connected abelian groups.

It should be possible to match up this $\mathbb{Z}_p$ with the $\mathbb{Z}_p$ that $u$ lives in above but I'm not entirely sure how.