Are there any non-contractible connected topological rings?
Of course, such a thing cannot be a (topological) algebra over the reals.
(I have a vague memory of having a glance at an erticle by Lurie in which some (for me) rather esoteric theory of higher categorical structures gave rise to topological rings that would have some very nontrivial topology, but I know nothing about that field(s) and, well, I just don't remember... Maybe someone can provide less "esoteric" examples! :) )
Here is a method for manufacturing such topological rings.
The main technical ingredient is a product-preserving functor
$$\Theta: \mathrm{Set}^{\Delta^{op}} \to \mathrm{CGHaus}$$
from the category of simplicial sets to the category of compactly generated Hausdorff spaces that is not, however, the usual geometric realization functor. This will almost undoubtedly be unfamiliar, and so will require some preface. The basic idea though is that while the usual geometric realization uses for its topological input the usual interval $I = [0, 1]$, the formal properties of the realization functor, particularly the fact it preserves finite products, still hold upon replacing $I$ by any compact topological interval $L$ and replacing ordinary affine simplices by $L$-valued simplices. This $L$-based realization $\Theta$, being product-preserving, takes simplicial rings to ring objects in $\mathrm{CGHaus}$. By choosing an appropriate $L$ that is connected but not path-connected, we can construct a topological ring that is connected but not path-connected, hence not contractible.
We define an interval to be a linearly ordered set with distinct top and bottom elements, and an interval map as an order-preserving map that preserves the top and bottom. Observe that the usual affine simplex $\sigma_{n-1}$ of dimension $n-1$ can be described as the space of $(n-1)$-tuples $0 \leq x_1 \leq \ldots \leq x_{n-1} \leq 1$ (topologized as a subspace of $I^n$), or in other words as the space of interval maps $[n+1] \to I$ from the finite interval with $n+1$ points to $I$. Meanwhile, the category $\mathrm{FinInt}$ of finite intervals $[n+1]$ is equivalent to $\Delta^{op}$ (where $\Delta$ is the category of finite nonempty ordinals); indeed we have a functor $\hom(-, [2]): \Delta^{op} \to \mathrm{FinInt}$, where the set of order-preserving maps $\hom([n], [2])$ from the $n$-element ordinal $[n]$ to $[2]$ is given the pointwise order, thus inheriting an interval structure from the interval structure on $[2]$, where we have $[n+1] \cong \hom_{\Delta}([n], [2])$ as intervals).
The usual geometric realization $R(X)$ of a simplicial set $X$, from a categorical point of view, is a tensor product $X \otimes_\Delta \sigma$ of a "right $\Delta$-module" $X: \Delta^{op} \to \mathrm{Set}$ with a left $\Delta$-module $\sigma: \Delta \to \mathrm{CGHaus}$ (the affine simplex functor):
$$\sigma: \Delta \simeq \mathrm{FinInt}^{op} \stackrel{\hom(-, I)}{\to} \mathrm{CGHaus}$$
$$[n] \mapsto [n+1] \mapsto \hom_{\mathrm{Int}}([n+1], I).$$
This tensor product is often described by a coend formula
$$R(X) = X \otimes_\Delta \sigma = \int^{[n] \in \Delta} X([n]) \cdot \hom_{\mathrm{Int}}([n+1], I).$$
As is well-known, $R$ is product-preserving. What is perhaps less well-known is that the only thing we need from $I$ to prove this fact is that it's compact Hausdorff and the interval order $\leq$ is a closed subset of $I \times I$. Complete details may be found in the nLab here. Therefore, if we replace $I$ with another compact Hausdorff topological interval $L$ (so that $\leq_L$ is a closed subset of $L \times L$), we get the same result, that the functor $\Theta = R_L$ defined by the formula
$$R_L(X) = \int^{[n] \in \Delta} X([n]) \cdot \hom_{\mathrm{Int}}([n+1], L)$$
is also product-preserving.
Let us take our compact topological interval $L$ to be the end-compactification of the long line (so, adjoin points $-\infty$ and $\infty$ to the ends of the long line). This is connected, but not path-connected because for example there is no path from $\infty$ to any other point. Now we just turn a crank: start with any denumerable non-trivial ring $R$ in $\mathrm{Set}$ -- I'll take $R = \mathbb{Z}/(2)$ -- and apply a sequence of product-preserving functors,
$$\mathrm{Set} \stackrel{K}{\to} \mathrm{Cat} \stackrel{N}{\to} \mathrm{Set}^{\Delta^{op}} \stackrel{R_L}{\to} \mathrm{CGHaus}.$$
(Here $K$ is the functor that takes a set $S$ to the category such that $\hom(x, y)$ is a singleton for any $x, y \in S$; this is right adjoint to the forgetful functor $\mathrm{Cat} \to \mathrm{Set}$ that remembers only the set of objects, and being a right adjoint, $K$ preserves products. The nerve functor $N$ also preserves products.) Since ring objects can be defined in any category with finite products, we have that product-preserving functors transport ring objects to ring objects. One should draw a picture of the category $K(\mathbb{Z}/(2))$; it's pretty clearly connected, and its nerve will be a connected simplicial set, or indeed a connected simplicial ring. The $L$-based realization of that will thus be a connected colimit of (connected) $L$-based simplices $\sigma_L(n) = \hom([n+1], L)$ (see the nLab here for connected colimits of connected spaces), and so it too will be a connected ring object in $\mathrm{CGHaus}$.
At this point, the overall idea should be pretty clear, and the rest is just some technical mopping-up.
One technical point is that products in $\mathrm{CGHaus}$ need not be usual topological products (as shown by a famous example of Dowker), so one might object that we could end up not with a topological ring, but some kind of funny ring object in $\mathrm{CGHaus}$. However, in many cases of interest, topological products do coincide with $\mathrm{CGHaus}$ products. This is particularly the case for colimits of countable increasing sequences of compact Hausdorff spaces: their product in $\mathrm{CGHaus}$ is the usual topological product. (The same proof as given by Allen Hatcher for Theorem A.6 here will do.) Thus, what counts here is that $N(K(\mathbb{Z}/(2)))$ is a simplicial set with finitely many cells in each dimension, and $R_L$ applied to this involves taking a countable union of compact Hausdorff spaces, so we are okay here.
A second technical point involves showing that $X = (R_L \circ N \circ K)(\mathbb{Z}/(2))$ is not path-connected, which is intuitively clear, but an idea of proof would be nice. $X$ can be described as a union of nondegenerate simplices, where there are two such simplices in each dimension $n$ (corresponding to paths of length $n$ of the form $0 \to 1 \to 0 \to \ldots$ and $1 \to 0 \to 1 \to \ldots$), and a point in the interior of each such simplex has coordinates given by an increasing chain of length $n$ in a dictionary order, say $(j_1, t_1) < (j_2, t_2) < \ldots < (j_n, t_n)$ where the $j_k$ belong to the order type $-\omega_1 \cup \omega_1$ ($\omega_1$ being the first uncountable ordinal, and $-\omega_1$ is of opposite order type, extending in the "negative" direction), and the $t_k$ belong to $[0, 1)$. Every point of $X$ is an interior point of some unique $n$-simplex. Now if $\alpha: I \to X$ is a path connecting a point in the interior of an $n$-simplex, $n > 0$, to a 0-simplex, then let $(a, b) \subset I$ be a connected component of the open set of $t \in I$ such that $\alpha(t)$ is interior to an $n$-simplex with $n > 0$. Since $(a, b)$ has countable cofinality, there is a countable ordinal $\kappa$ such that for every $t \in (a, b)$, the maximum ordinal $|j_k|$ occurring in the coordinate description of $\alpha(t)$ is bounded above by $\kappa$. But $\alpha(a)$, being a 0-cell, has a neighborhood $U$ where every point $p \in U$, $p \neq \alpha(a)$, has a maximum $|j_k|$ (in its coordinate description) greater than $\kappa$, and we have reached a contradiction.
If you replace "connected" with "path-connected", then no.
If 1 is in the same path component as 0, then choose such a path $\gamma$. Then the map $(x,t) \mapsto x \cdot \gamma(t)$ is a contraction of the space.
As a result, any counterexample would need to be connected but not locally path-connected. I do not know immediate examples of such a ring.
