I think there are a few things to untangle here.
First, as concerns your highlighted question, it seems that you've answered it yourself: outside the compact Hausdorff case (where the uniform structure is completely equivalent to the topology), it's unreasonable to think that the associated condensed set "knows" the uniform structure. It really only knows the topology.
Second, you are correct in your unpacking of the definition of "solid" as it applies to (let's say metrizable) toplogical abelian groups $M$. If $\underline{M}$ is solid, then for any nullsequence $(m_n)$ in $M$ and any sequence of integers $(a_n)$ the sum $\sum a_nm_n$ must converge in $M$. I agree that this feels a lot saying that $M$ is nonarchimedean and complete. However, though I gather for you "complete" means "Cauchy-complete" as usual, I'm not sure what general definition of a topological abelian group $M$ being "nonarchimedean" you're thinking of here to say that you think it is actually equivalent.
Here's something I find helpful to keep in mind. There are several differences between the notion of "solid" and the notion of "complete", beyond the fact that solid enforces some kind of non-archimedeanness. (Indeed, the same remarks apply in the liquid setting, which does not enforce nonarchimedeanness.)
First, while both the definition of "complete" and the definition of "solid" are of the form "for every [...] there exists a unique {...}", the differences in nature of the ...'s occuring lead to drastic divergence of the general notions. Most importantly, the solid condition in no way implies any kind of Hausdorff behavior. Essentially, the reason is that in the solid condition you already require in [...] a bunch of limits to exist uniquely (because you map in from a compact Hausdorff space). Then the solid condition is only that this generates more limits (and uniquely) by taking certain $\mathbb{Z}$-linear combinations. Whereas in the usual Cauchy completeness every limit that you posit exists, you also posit exists uniquely.
This phenomenon, that solid abelian groups incorporate non-Hausdorff behvaior, it absolutely crucial to having a functioning theory. Why? Because non-Hausdorff behavior is inevitable once you require an abelian category. If $M$ is some non-discrete condensed abelian group which you want to call "complete", and $N$ is any discrete dense subgroup mapping into $M$, the cokernel has to be "complete" as well, even though it's a classic example of a "bad quotient" which is non-Hausdorff.
(This is an example of the trade-off between "good categories of (mostly) bad objects" vs. "bad categories of good objects". But for me personally, I've seen enough examples of the solid formalism gracefully handling non-Hausdorff spaces in practice that I no longer think of them as "bad objects".)
You might be tempted to then think that solid abelian groups are more like weakened analog of complete topological abelian groups, where you drop the uniqueness requirement in the definition of completeness and only require existence of limits for Cauchy sequences. But no, the fact that solidness is an "exists a unique" condition is also crucial for it being an abelian category. "Exists" is just not stable enough a notion.
The other big difference is that a solid abelian group is "only required to be complete as far as compact subsets are concerned". If you have a Cauchy sequence which is not contained in a compact subset, the solid condition says nothing about it. Thus, for example, solid abelian groups are closed under arbitrary direct sums, whereas I don't think there's any reasonable topology on the direct sum $\oplus \mathbb{Z}_p$ for which it's complete [Edit: I thought wrong! See https://mathoverflow.net/questions/387322/countable-sum-bigoplus-n-0-infty-mathbb-z-p-as-a-topological-group]. Again, the fact that solid abelian groups are closed under all colimits is very important for us theoretically: it's part of what makes sure that the category of solid abelian groups "behaves like the category of modules over a ring" (formally, it is an abelian category generated by compact projective objects), and therefore has convenient algebraic properties.
Now, it seems the main thrust of your question as about defining possibile non-abelian analogs of solidness. I don't want to say this can't be done (I doubt it can but I certainly could be wrong), but I hope that the above remarks show that if you want to define such a notion, you shouldn't do it by trying to follow the usual presentation via Cauchy-completeness and uniform structures. Despite the fact that the two notions agree on many familiar objects, there is a huge divergence in general.
Not quite a complete answer, but:
There are models of constructive mathematics where sets are $T_0$ second-countable (optionally zero-dimensional) topological spaces equipped with equivalence relations, morphisms are continuous maps which respect the equivalence relation, and pointwise-equivalent morphisms are equal. In these two models, quotients are taken simply by inflating the equivalence relation, and the result satisfies all of the usual rules of MLTT + Quotient Types. (We need to add an equivalence relation structure so that the quotients are well-behaved enough for exponents to exist.)
In particular, each functions $f : 2^{\mathbb{N}}/\operatorname{fin} \to 2^{\mathbb{N}}/\operatorname{fin}$ arises from some continuous function $g:2^{\mathbb{N}} \to 2^{\mathbb{N}}$
These models don't have a set of truth values (hence no power-set operation), but they can be upgraded to contain one. (Roughly speaking - the above models arise as categories of PERs over two different PCAs, and there's a general technique for upgrading these, called realisability topoi.) This is the smallest change I know to the category of topological spaces that results in a model of MLTT.
As for what "every function is continuous" means - there are three ways that I know of. The first is, to every set, assign a topology (we could say that the opens are the unions of preimages of arbitrary functions into $\mathbb{R}$, for instance), and then every function is continuous with respect to this topology. This could also be done in classical mathematics, but there all sets would be given the discrete topology. Constructively, it may be the case that we get nontrivial topologies, and may even happen to get the 'correct' topologies for spaces such as $2^{\mathbb{N}}$ or $\mathbb R$.
The second way gives a number of examples where this holds. It is consistent that all functions $2^{\mathbb{N}} \to \mathbb{R}$ are continuous. In the presence of countable choice (which is consistent with this; iirc both (and MP, mentioned below) hold in Johnstone's topological topos, which has FirstCountableTop as a full subcategory and which I suspect is a full subcategory of that of condensed sets), this implies the stronger theorem that If $M$, $N$ are metric spaces with $M$ complete, then all functions $f:M \to N$ are continuous. (If MP ($\not \geq_{\mathbb{R}}$ implies $<_{\mathbb{R}}$) holds, we can weaken "complete" to "complement of a subset of a complete metric space".)
The third way is to generalise to grothendieck topologies, which is the main way to make (grothendieck) topoi, which model MLTT + quotients. Here, we specify a small category of "test spaces" and specify some families of morphisms to be "coverings" (for example, we could take all compact second-countable spaces for test spaces, and finite joint surjections as coverings) satisfying some technical conditions. The resulting grothendieck topos includes the spaces from the test category, and doesn't add any new morphisms between them - so if all of your morphisms between test spaces were continuous, they will continue to all be in the larger topos. The condensed sets are formed very similarly to this, except that the category of test spaces is no longer small, so we need to deal with size issues to get a grothendieck topos, like Simon mentioned.
Best Answer
Great question!
The answer is Yes. Let me elaborate a little. The question is more generally about the left adjoint to the inclusion $\mathrm{An}\to \mathrm{CondAn}$ from anima to condensed anima. This left adjoint is only partially defined; what exists in general is the functor $F: \mathrm{CondAn}\to \mathrm{Pro}(\mathrm{An})$ from condensed anima to pro-anima. Then $F(X)$ is an anima precisely when the left adjoint to $\mathrm{An}\to \mathrm{CondAn}$ exists on $X$, in which case it equals $F(X)$.
In other words, for any condensed anima $X$ and any anima $Y$, the map $$ \mathrm{Hom}(X,Y)\to \mathrm{Hom}(X\times [0,1],Y) $$ is an isomorphism. Writing $X$ as a colimit of profinite sets $S$, this reduces to the case $X=S$. But then the condensed anima $\mathrm{Hom}(S,Y)$ is actually itself an anima (if $S=\mathrm{lim}_i S_i$, it is the colimit $\mathrm{colim}_i \mathrm{Hom}(S_i,Y)$), and by adjunction one reduces to the case that $X=\ast$. In this case, this is part of Lemma 11.9 in Analytic Geometry.
The lemma shows that $F$ factors over a functor $$\mathrm{CondAn}[W^{-1}]\to \mathrm{Pro}(\mathrm{An})$$ from the $\infty$-category obtained from $\mathrm{CondAn}$ by inverting homotopy equivalences.
In particular, any condensed anima $X$ which is homotopy equivalent to an anima $Y$ has the property that $F(X)=Y$. In particular, if $X$ comes from a contractible topological space, then $F(X)=\ast$.
Now if $X$ is just locally contractible, then one can find a cover $X=\bigcup_i U_i$ by contractible $U_i$, and then covers $U_i\cap U_j=\bigcup_k U_{ijk}$, etc., leading to a hypercover of $X$ by disjoint unions of contractible $U\subset X$. As $F$ commutes with colimits, this writes $F(X)$ as a colimit of disjoint unions of points, and hence $F(X)$ is an anima (which is the usual (weak) homotopy type).