There are infinitely many embeddings. However, all but one of them is "essentially the same as" the one you studied as they become equal to the one you studied on restriction to $SU(2)\times SU(3)$. The remaining one is the one studied by Krasnov.
I follow the strategy suggested by Kenta Suzuki.
$SU(3)$ has irreducible representations of dimensions $1,3,3,6,8,6, 10, 10$, and higher dimensions. The $10$-dimensional ones are dual to each other, as are the $6$-dimensional ones, so they can't appear. The $3$-dimensional ones are dual to each other and can only appear together. So the only $10$-dimensional self-dual representations of $SU(3)$ decompose as irreducibles as $8+1+1$, $3+3+1+1+1+1$, or ten $1$s. All of these are orthogonal because the 8-dimensional representation is orthogonal. However, the ten $1$s cannot appear because then $SU(3)$ would act trivially.
A representation of $SU(3) \times SU(2)$ is a sum of tensor products of irreducible representations of $SU(3)$ and irreducible representations of $SU(2)$. Restricted to $SU(3)$, each tensor product splits into a sum of copies of the same irreducible representation. So $SU(2)$ can only act nontrivially when the same representation appears multiple times. Since the $3+3$ is two different $3$-dimensional representation, only the $1$-dimensional representation can occur twice. Thus, our 10-dimensional orthogonal representation of $SU(3) \times SU(2)$ necessarily splits as either the $8$-dimensional adjoint repsentation of $SU(3)$ plus a $2$-dimensional orthogonal representation of $SU(2)$ or the $6$-dimensional sum of standard and conjugate representations of $SU(3)$ plus a $4$-dimensional orthogonal representation of $SU(2)$. However, $SU(2)$ has a unique nontrivial representation of dimension $2$ and it isn't orthgonal, so only the second case can appear. $SU(2)$ has representations of dimension $1,2,3,4$ of which the $2$ and $4$-dimensional ones are symplectic and so must appear with even multiplicity in any orthogonal representation, so the only nontrivial $4$-dimensional orthogonal ones are $2+2$ or $3+1$.
So there are two ten-dimensional orthogonal representations of $SU(2) \times SU(3)$ that are nontrivial on both factors, those being the sum of two different $3$-dimensional irreducible representations of $SU(3)$ with either two copies of the two-dimensional irreducible representation of $SU(2)$ or the three-dimensional and the one-dimensional irreducible representation of $SU(2)$. The orthogonal structure is unique up to isomorphisms, so these give two conjugacy classes of homomorphisms $SU(2) \times SU(3) \to SO(10)$ and thus two conjugacy classes of homomorphisms $SU(2) \times SU(3) \to \operatorname{Spin}(10)$. The first one corrresponds to the embedding you studied while only the second one restricts to $\operatorname{Spin}(9)$ so indeed these are different.
To understand how to extend these to $SU(2) \times SU(3)$, I consider the centralizer of the representation within $\operatorname{Spin}(10)$. Since the group is connected, this is the same as the centralizer of its Lie algebra, which is therefore the inverse image of the centralizer in $\operatorname{SO}(10)$. Now there is a distinction between the two examples because the example with irrep dimensions $3+3+2+2$ has centralizer with identity component $U(1) \times SU(2)$ while the example with irrep dimensions $3+3+3+1$ has centralizer with identity component $U(1)$. In the second case, the image of $U(2) \times U(3)$ must be the image of $SU(2) \times SU(3)$ times the centralizer of the image of $SU(2) \times SU(3)$, so this gives a unique example, which must be the one considered by Krasnov.
In the first case, we can restrict attention to a torus $U(1) \times U(1)$ in $SU(2) \times SU(2)$. The center of $S(U(2) \times U(3))$ maps to a one-dimensional subgroup of this torus, which can be described by a pair of integers. Explicitly, given a two-by-two-unitary matrix $A$ and a three-by-three unitary matrix $B$ with $\det(A) \det(B) =1$, we can map to $U(5)$ by sending $(A,B)$ to $A \gamma^a \oplus B \gamma^b$ where $\gamma = \det (A) = \det(B)^{-1}$, and then map from $U(5)$ to $SO(10)$. This lifts to the spin group if and only if the determinant in $U(5)$ is a perfect square. The determinant is $\gamma^{ 1 + 2a - 1 + 3b} = \gamma^{2a+3b}$ so a lift exists if and only if $b$ is even.
The only possible kernel of this embedding is the scalars. The scalar $A = \lambda^3 I_2, B = \lambda^{-2} I_3$ maps to $\lambda^{3+ 6a} I_2 \oplus \lambda^{-2 + 6b} I_3$ and so the kernel is trivial if and only if $\gcd(3+6a,-2 + 6b)=1$.
However, there are infinitely many integer solutions $a,b$ to $\gcd(3+6a,-2a+6b)=1$ with $b$ even (in fact, a random $a$ and even $b$ works with probability $9/\pi^2$), so this gives infinitely many examples.
