Mathematical logic (specifically model theory) provides a partial answer. Let $M$ and $N$ be structures for a first-order language $L$. $M$ and $N$ are elementarily equivalent if every closed formula satisfied by one is satisfied by the other. $M$ and $N$ are isomorphic if there is a 1-1 map between $M$ and $N$ that preserves all the relations and functions mentioned in the signature of $L$. Theorem: if $M$ and $N$ are isomorphic, then they're elementarily equivalent. See, say Marker Model Theory: An introduction, §1.1, or Hodges A Shorter Model Theory, §1.2.
I think this serves as a reasonable candidate for "a general theorem that all such properties/objects are preserved by isomorphisms in the category we're working in".
I say a partial answer, because choosing the language in each case remains an issue. Let me elaborate for your example of groups. We want to show that being a subgroup, or a normal subgroup, or the center, is preserved by isomorphisms, all in one shot. For $L$, we include the following in its signature: the constant symbol 1, the function symbols $\cdot,{}^{-1}$, and a unary relation symbol $S$ for the subset under discussion. (There are other signatures that would also serve.) Here are the closed formulas that express "$S$ is a subgroup", etc. I'm going to be a bit sloppy for increased readability, using juxtaposition for the operation and omitting parentheses. Also, when I write "$S$ is a subgroup" in the second two bullets, just imagine the first bullet being repeated in full.
- $S(1)\wedge\forall x\forall y[S(x)\wedge S(y)\rightarrow S(x^{-1})\wedge S(xy)]$
- $S$ is a subgroup and $\forall x\forall y[S(x)\rightarrow S(y^{-1}xy)]$
- $S$ is a subgroup and $\forall x[\forall y(yx=xy)\rightarrow S(x)]\wedge
\forall x[S(x)\rightarrow\forall y(yx=xy)]$
So if $M$ and $N$ are isomorphic, then $M$ satisfies one of these formulas if and only if $N$ does—that's what elementary equivalence says. And if $M$ and $N$ are isomorphic groups, then the subsets defined by the relation symbol $S$ correspond, and therefore one is a subgroup (or normal, or the center, or anything expressible by a closed formula in this language) if and only if the other is.
If you're familiar with first-order logic, you'll be aware of various hurdles to overcome. For example, to define "commutator subgroup" with a closed formula, you'd need to expand the language to allow for sequences of arbitrary finite length, since the commutator subgroup is generated by the commutators. That means incorporating $\mathbb{N}$ into the structure in some manner. I don't mean that $\mathbb{N}$ would be a subset of the group, rather that the structure would be (implicitly) an ordered tuple $(G,\mathbb{N},\ldots)$. For "derived series" you'd need to expand the language some more. But all these obstacles can be mastered with standard techniques.
A fuller answer would discuss connecting the category theory with the model theory. I plead limitations of both space and my expertise.
This answer may not be in the spirit of the question, but I think it's worth noting.
The usual definition of Hausdorff uses the notions "point" and "open sets". But both of these notions can be expressed in the language of categories: a point of $X$ is an arrow from the one-point space $1\to X$, and an open set of $X$ is an arrow from $X$ to the Sierpiński space $X\to S$. So you can translate the usual definition to the language of categories.
Recall that $S=\{0,1\}$ with open sets $\varnothing$, $\{1\}$, and $S$. Overloading notation, write $1$ for the arrow $1\to S$ mapping the unique point of $1$ to $1$.
$X$ is Hausdorff if and only if: for every pair of arrows $p_1\neq p_2\colon 1 \to X$, there is a pair of arrows $u_1,u_2\colon X\to S$ such that (1) $u_1\circ p_1 = u_2\circ p_2 = 1$, and (2) there is no arrow $q\colon 1\to X$ such that $u_1\circ q=u_2\circ q=1$.
Best Answer
A frame $F$ is spatial if and only if for all $U,V\in F$, if $U\not\leq V$, then there exists a completely prime filter $p$ on $F$ such that $U\in p$ and $V\notin p$.
Note that this is characterization 3 on the nLab page you linked to, remembering that a "point" $p$ of $F$ is a completely prime filter on $F$ (equivalently, a frame homomorphism $F\to 2$), and a point $p$ "belongs to" $U$ if and only if $U\in p$.
Proof: Suppose $F$ is spatial. Then there is a topological space $X$ and an isomorphism $i\colon F\cong \mathrm{Open}(X)$, the frame of open sets in $X$. Suppose $U\not\leq V$ in $F$. Then $i(U)\not\subseteq i(V)$, so there is a point $x\in i(U)\setminus i(V)$. Let $p_x = \{W\in F\mid x\in i(W)\}$. Check that $p_x$ is a completely prime filter in $F$ such that $U\in p_x$ and $V\notin p_x$.
Conversely, suppose that for all $U,V\in F$, if $U\not\leq V$, then there exists a completely prime filter $p$ on $F$ such that $U\in p$ and $V\notin p$. Let $X$ be the space of points of $F$, whose points are completely prime filters in $F$ and whose open sets are of the form $[U] = \{p\in X\mid U\in p\}$ for $U\in F$. Check that these sets form a topology on $X$ and the map $i\colon U\mapsto [U]$ is a frame homomorphism $F\to \mathrm{Open}(X)$. This map is surjective by definition of the topology on $X$, so it remains to check that it is injective. If $U\neq V$ in $F$, then without loss of generality $U\not\leq V$. Let $p$ be a completely prime filter such that $U\in p$ and $V\notin p$. Then $p$ is a point of $X$ witnessing that $[U]\neq [V]$, so $i$ is an isomorphism of frames.