This falls slightly short of a proof but is too long for a comment.
The question inquires into the largest family of triangulations in which every pair of members shares some diagonal.
If we instead fix any single diagonal in common to every member of a family of triangulation of an n-gon, it is self-evident (from the noncrossing property of triangulations) that the number of triangulations sharing it is given by the product of the enumerations of the two families of triangulations of the ploygons so formed either side of it.
For any triangulatable n-gon, any given diagonal is a member of up to two fan triangulations, in which it can be assigned an integer the $d^{th}$ diagonal from the perimeter satisfying $1\leq d\leq \frac{n-2}{2}$. By symmetry we may choose either of the two fans with no effect upon the distance of the diagonal from the nearest edge.
$d=\frac{n-2}{2}$ shall be the distance of the central triangulation if $n$ is even and $d=\frac{n-3}{2}$ shall be the distance of the two "most central" diagonalisations if $n$ is odd.
The number of triangulations of any $n$-gon sharing the $d^{th}$ diagonal is the product of the number of triangulations $\cal T(d,n)$ either side of it:
$$\cal T(d,n)=C_d\times C_{n-d}$$
$$\cal T(d,n)=(d+1)^{-1}\binom{2d}{d}(n-d+1)^{-1}\binom{2n-2d}{n-d}$$
The ratio of one Catalan number to the next is greater than the ratio of the previous two Catalan numbers, and therefore in the choice of which diagonal to share, in order to maximise the size of the family, for any given choice $d$ of diagonal which is not at the edge, we may always identify a larger family by choosing the diagonal $d-1$ which is one position closer to the perimeter and by induction the outermost diagonal $d=1$ yields the largest family.
The size of the largest family of triangulations sharing the same diagonal is therefore given by:
$|\cal S| = C_{n-2-1}\times C_1=|\cal T_{n-1}|$
My claim which would complete the proof (and which remains to be proven) is that there is no family of triangulations in which every pair shares some diagonal, which is larger than the largest family of triangulations in which every triangulation in the family shares the same diagonal.
I suspect this is proven by the fact that the largest face on any associahedron has the same number of edges as the number of vertices of the associahedron one order smaller. In particular I believe that each face of the associahedron represents a family of triangulations which all share the same edge. These two statements alone would in fact be sufficient to answer the entire question in the affirmative.
I think this is indeed true. This should follow from the rigidity of convex spherical polytopes and a few more lemmas. I'll explain how to deal with cases $d\ge 3$.
Lemma 1, rigidity. Two convex spherical cobminatorially equivalent polytopes with isomertric faces are isometric.
Lemma 2, bounded volume. The boundary of any convex subset of round $\mathbb S^n$ has volume bounded from above by the volume of round $\mathbb S^{n-1}$.
Lemma 3, gap. Suppose we have a spherical convex polytope in $S^n$ of diameter less than $\pi-\varepsilon$ then the volume of the dual polytope is bounded from below by $c(n,\varepsilon)>0$.
Lemma 4, sum of dual solid angles. Let $P$ be a convex $d+1$-polytope. For each of its vertices consider the dual cone. Then the sum of solid angles of such dual cones is equal to the volume of the round $\mathbb S^d$.
Proof of finitness. So, suppose we have a finite number of convex $d$-polytopes $P_1,\ldots P_k$. Denote by $S_1,\ldots, S_m$ the spherical $d-1$ polytopes corresponding to vertices of $P_1,\ldots, P_k$.
Suppose we have a $d+1$-polytope $P$ with faces homothetic to $P_1,\ldots P_k$. Then for each vertex of $P$ we get a convex spherical $d$-polytope whose faces are composed from $S_1,\ldots, S_m$. Note the the number of such spherical polytopes is finite up to isometry, by Lemmas 1 and 2. Moreover, all of them have diameter less than $\pi-\varepsilon$ for some $\varepsilon$ . So by Lemma 3 the volumes of spherical polytopes dual to those (coming from vertices) are at least $c(d,\varepsilon)$. Hence by Lemma 4 the total number of vertices of $P$ is at most $vol(S^d)/c(d,\varepsilon)$. Hence we have only finite number of combinatorial types of $P$.
Best Answer
There are other polytopes with this property that can be obtained via the free join construction.
Given two polytopes $P_1\subset\Bbb R^{d_1}$ and $P_2\subset\Bbb R^{d_2}$, the free join $P_1\bowtie P_2$ is obtained by embedding $P_1$ and $P_2$ into skew affine subspaces of $\Bbb R^{d_1+d_2+1}$ and taking the convex hull.
The claim is that if $P_1$ and $P_2$ have your property, then so does $P_1\bowtie P_2$. To see this, note that the facets of $P_1\bowtie P_2$ are exactly of the form $P_1\bowtie f_2$ and $f_1\bowtie P_2$, where $f_i$ is a facet of $P_i$. Now, if $v$ is a vertex in, say, $P_1\subset P_1\bowtie P_2$ that is in all facets of $P_1$ except for $f,f'\subset P_1$, then $v$ is in all facets of $P_1\bowtie P_2$ except for $f\bowtie P_2$ and $f'\bowtie P_2$. Equivalently for vertices in $P_2$.
Example. Take the free join of two squares, which is a 5-dimensional polytope with 8 vertices, 8 facets and vertex degree 6. This polytope is not simple and can thus not be the cartesian product of two simplices.
This construction yields counterexamples in dimension $d\ge 5$. Concerning $d=4$, there should not be any counterexamples in this dimension.
Suppose that $P\subset\Bbb R^4$ is a counterexample. According to Moritz Firsching's comment we can assume that it has $n\ge 10$ facets. Each vertex is then contained in exactly $n-2\ge 8$ of them. More generally, each set of $k$ vertices is contained in at least $n-2k$ facets. But let $f\subset P$ be a 2-face of $P$ and $v_1,v_2,v_3\in f$ three affinely independent vertices, then this set of vertices, and thus also $f$, is contained in $n-6\ge 4$ facets. This cannot be, since each 2-face of $P$ is contained in exactly two facets.
I believe this generalizes to show that any polytope with your property must have $\le 2d$ facets.