Going in the opposite direction than stable theories, you can use Gödel’s theorems to prove finite nonaxiomatibility:
Theorem: If $T$ is a consistent complete theory interpreting Robinson’s arithmetic $\mathrm Q$, then $T$ is not finitely axiomatizable.
(In fact, not even recursively axiomatizable.)
Theorem: If $T$ is a consistent reflexive theory, then $T$ is not finitely axiomatizable.
Here, a theory is reflexive if there is a fixed interpretation $I$ of $\mathrm Q$ in $T$ such that $T$ proves $\mathrm{Con}_S^I$ for every finitely axiomatized subtheory $S$ of $T$.
A useful criterion, mentioned in https://mathoverflow.net/questions/87249, is
Theorem: Any sequential theory proving the induction schema for all formulas in its language is reflexive (hence not finitely axiomatizable unless inconsistent).
(A theory is sequential if it can define a coding of finite sequences of its objects, loosely speaking.) In particular, all consistent extensions (in the same language) of Peano arithmetic or of set theories like ZF are non-finitely axiomatizable.
$\newcommand{\o}[0]{\mathsf{orth}}$No, these equations do not yield the complete theory of the orthocenter.
The identity
$$\o(\o(t,u,v),\o(t,u,w),u) = \o(\o(t,u,v),\o(t,v,w),v)$$
holds for the orthocenter (X(4)) but not for X(74) (the isogonal conjugate of the Euler infinity point), even though both satisfy involutory identities like
$$v=\o(t,u,\o(t,u,v))$$
I found the first identity and tested both using Mathematica, which can set up the functions quickly as follows:
avg[a_, b_, c_, u_, v_, w_] := (a u + b v + c w)/(a + b + c);
bary[f_, u_, v_, w_] := avg[f[u,v,w], f[v,w,u], f[w,u,v], u, v, w];
cosA[u_, v_, w_] := (v-u).(w-u) / Sqrt[((v-u).(v-u)) ((w-u).(w-u))];
f[u_, v_, w_] := 1/((u-v).(u-w));
g[u_, v_, w_] := Sqrt[(v-w).(v-w)] / (cosA[u,v,w] - 2 cosA[v,w,u] cosA[w,u,v]);
center4[u_, v_, w_] := bary[f, u, v, w];
center74[u_, v_, w_] := bary[g, u, v, w];
Then the following code tests the identity for the orthocenter algebraically, and for the X(74) center numerically:
Algebra = {t -> {tx, ty}, u -> {ux, uy}, v -> {vx, vy}, w -> {wx, wy}};
Example = {t -> {0, 1}, u -> {2, 4}, v -> {-3, 5}, w -> {-2, 1}};
FourVariable[c_] := c[c[t,u,v], c[t,u,w], u] == c[c[t,u,v], c[t,v,w], v];
{FourVariable[center4] /. Algebra // Simplify, FourVariable[center74] /. Example}
and the tests return True and False respectively.
Best Answer
This algebra is finitely based.
In fact, if you choose any bijection from $\{a,b,c,d\}$ to $\mathbb Z_2\times \mathbb Z_2$, then you can transport the operation $F(x,y,z)$ to $\mathbb Z_2\times \mathbb Z_2$ it find that it is $x-y+z$. The resulting algebra $\langle \mathbb Z_2\times \mathbb Z_2; x-y+z\rangle$ is the square of the algebra $\langle \mathbb Z_2; x-y+z\rangle$. Since an algebra and its square have the same equational theory, the question reduces to: Is $\langle \mathbb Z_2; x-y+z\rangle$ finitely based?
Roger Lyndon proved in 1951 that all $2$-element algebras are finitely based. In his paper, the algebra $\langle \mathbb Z_2; x-y+z\rangle$ is referred to as 'system $L_4$'. This is one of fifteen systems he deals with in Section I of his paper. Lyndon does not provide an equational basis for this algebra, but writes ``a complete set of axioms [for each of the systems in Section I] can be chosen by inspection from the various familiar sets of axioms for Boolean algebras and Boolean rings''.
I claim that the following identities suffice:
The reason these are enough is that $F(x,x,y)=F(y,x,x)=y$ (part of the first bullet point) states that $F$ is a Maltsev operation. The law from the second bullet point states that $F$ commutes with itself. H. P. Gumm showed that any Maltsev operation that commutes with itself on a set $A$ must be $x-y+z$ for some abelian group structure on $A$. Now the law $F(x,y,x) = y$ implies that $-y=y$ for this abelian group. Hence any model of the bulleted identities must be $\langle A; F(x,y,z)\rangle$ where $F(x,y,z) = x-y+z$ for some abelian group of exponent $2$ on $A$. Any two such things have the same equational theory. In fact, if $\langle A; F(x,y,z)\rangle$ and $\langle B; F(x,y,z)\rangle$ are nontrivial and satisfy these identities, then each is embeddable in a power of the other. This is enough to prove that the set of bulleted identities axiomatizes an equationally complete theory, hence forms a basis of identities for any of its nontrivial models.