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I concur with your comments -- it does seem like the words "finitely presented" should be included in the definition, even if perhaps they are implied (very non-obviously!) by the other conditions. Thanks for the comments, this has really cleared this up for me!
A small variant on the strategy @WillSawin proposes: if $f$ is a modular function of weight $0$, holomorphic away from the cusp and with a simple pole at the cusp, then there is some complex number $a$ such that $f-aj$ is holomorphic at the cusp (cancel out the leading term of the pole). The same Liouville argument shows that $f-aj$ is constant, i.e. $f=aj+b$. If you want your normalisation conditions $f(e^{2\pi i/3})=0$ and $f(i)=1$, then this forces $a=1$ and $b=0$, so $f=j$.
@AchimKrause By way of analogy, consider the "free abelian group" functor, from {sets} to {abelian groups}. The free abelian group $F_{ab}(S)$ on a set $S$ is characterised by the fact that $\mathrm{Hom}(F_{ab}(S),G)\to\mathrm{Map}(S,G)$ is an isomorphism for all abelian groups $G$. The free group $F(S)$ on $S$ also satisfies this same universal property, but this does not imply that $F(S)=F_{ab}(S)$ in general, exactly because $F(S)$ can be non-abelian. I could envisage something similar happening with strict vs. non-strict Picard groupoids.
@AchimKrause I'm not sure I follow the logic that goes from "the free Picard groupoid on $S$ is not strict" to "there is no free strict Picard groupoid on $S$". That is, because I'm requiring the universal property only for $\mathcal G$ strict, it is not automatic that a free strict Picard groupoid would be a free Picard groupoid, or vice versa. (I wasn't clear in the question that $\mathcal G$ should be assumed strict; I'll clarify that now). Put another way, I believe that $\Omega^\infty\Sigma^2H\mathbb Z$ is not strict, so doesn't give a counterexample to $\mathbb Z$ being free strict.
@AchimKrause By a Picard groupoid, I mean a symmetric monoidal 1-category in which all morphisms are invertible and the functor $X\otimes(-)\colon \mathcal G\to\mathcal G$ is an equivalence for all objects $X$. By a strict Picard groupoid, I mean a Picard groupoid in which the commutativity constraint $\beta_{X,X}\colon X\otimes X \to X\otimes X$ is the identity for all objects $X$. These are sometimes called strictly commutative Picard groupoids.
By the way, the map $G\to U(G)(\mathbb Q_p)$ in the first part, and the map $G\to U(H)(\mathbb Q_p)$ in the second part exhibit this group $U(G)$ or $U(H)$ as the $\mathbb Q_p$-Malcev completion (or pro-unipotent completion) of the topological group $G$, meaning that it is the initial continuous group homomorphism from $G$ to the $\mathbb Q_p$-points of a pro-unipotent group.
@DavidESpeyer Actually the $p$-Sylow of $GL_n(\mathbb Z_p)$ is bigger. Even when $n=1$, the $p$-Sylow subgroup in $GL_1(\mathbb Z_p)=\mathbb Z_p^\times$ is $1+p\mathbb Z_p$. And in general a $p$-Sylow subgroup of $GL_n(\mathbb Z_p)$ is the preimage of $U_n(\mathbb F_p)$ under the map $GL_n(\mathbb Z_p) \twoheadrightarrow GL_n(\mathbb F_p)$, since the kernel of that map is a pro-$p$ group.
Ah, interesting! I'd not met this notion of linearly compact topological vector spaces before -- I was used to thinking of these kinds of Lie algebras inside the pro-category of finite-dimensional vector spaces. These two categories (pro-finite-dimensional vector spaces and linearly compact topological vector spaces) are equivalent, but the latter gives a more concrete way of thinking about the same objects.
Yes, this is why I find this result surprising: coassociative coalgebras are always locally finite, as are comodules under them. I'll have to think about whether the Lie algebra you mention could lead to an example. The correct "dual" notion to Lie coalgebras over $\mathbb{R}$ is topological Lie algebras whose underlying topological vector space is isomorphic to $\mathbb{R}^\kappa$ for some cardinal $\kappa$, taking the product topology of the usual topology on $\mathbb{R}$ (or the discrete topology, they give equivalent categories). Does your Lie algebra carry such a topology?
This is great -- in fact what I was originally looking for was phrased on the topological Lie algebra side anyway. Is the final part true in general? I.e. is any Lie coalgebra the union of its finite-dimensional Lie coideals?
