Indeed, there are no nonzero injective condensed abelian groups.
Let $I$ be an injective condensed abelian group. We can find some surjection
$$ \bigoplus_{j\in J} \mathbb Z[S_j]\to I$$
for some index set $J$ and some profinite sets $S_j$, where $\mathbb Z[S_j]$ is the free condensed abelian group on $S_j$ -- this is true for any condensed abelian group. But now we can find an injection
$$\bigoplus_{j\in J} \mathbb Z[S_j]\hookrightarrow K$$
into some compact abelian group $K$, for example a product of copies of $\mathbb Z_p$ for any chosen prime $p$. Indeed, it suffices to do this for any summand individually (embedding into a product in the end), and each factor embeds into a product of copies of $\mathbb Z$ (by choosing many maps $S\to \mathbb Z$), thus into a product of copies of $\mathbb Z_p$.
We remark that it is in this step that we need to work in the condensed setting: In the pyknotic setting, $J$ can be larger than the relevant cutoff cardinal for the profinite sets, so $K$ would not be in the site of $\kappa$-small compact Hausdorff spaces.
By injectivity of $I$, we get a surjection $K\to I$. In particular, the underlying condensed set of $I$ is quasicompact.
Now assume that $I$ is $\kappa$-condensed for some $\kappa$, and pick a set $A$ of cardinality bigger than $\kappa$, and consider the injection
$$\bigoplus_A I\hookrightarrow \prod_A I.$$
The sum map $\bigoplus_A I\to I$ extends to $\prod_A I\to I$ by injectivity of $I$.
I claim that the map $\prod_A I\to I$ necessarily factors over a map $\prod_{A'} I\to I$ for some subset $A'\subset A$ where the cardinality of $A'$ is less than $\kappa$. To check this, we use the surjection $K\to I$; then it is enough to prove that the map $\prod_A K\to I$ factors over $\prod_{A'} K\to I$ for some such $A'$. But this follows from $I$ being $\kappa$-condensed and $\prod_A K$ being profinite.
Thus, the sum map $\bigoplus_A I\to I$ factors over $\prod_{A'} I\to I$ for some $A'\subset A$. But then restricting the sum map along the inclusion $I\to \bigoplus_A I$ given by some $a\in A\setminus A'$ gives both the identity and the zero map, finally showing that $I=0$.
I hope I didn't screw something up.
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.
Best Answer
I'll ignore the set-theoretic issues since I don't understand them well enough to say anything about them.
As with any site, the limits (and in particular, the kernel) of sheaves may be computed pointwise. On the other hand, the colomits are usually not computed pointwise (cokernels included). For example, an exact sequence of condensed Abelian groups $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $ gives a long exact sequence of cohomology $$ 0 \rightarrow A(S) \rightarrow B(S) \rightarrow C(S) \rightarrow H^1(S,A) \rightarrow \dotsc $$
This means that the cokernels should be given by the pointwise quotient if the map $H^1(S,A) \rightarrow H^1(S,B)$ is injective (this happens, for example, when $H^1(S,A)=0$).