The category of von Neumann algebras W* admits a variety of monoidal structures of three distinct flavors.
(1) W* is complete and therefore you have a monoidal structure given by the categorical product.
(2a) W* is cocomplete and therefore you also have a monoidal structure given by the categorical coproduct.
(2b) I suspect that there is also a “spatial coproduct”, just as there is a categorical
tensor product and a spatial tensor product (see below).
The spatial coproduct should correspond to a certain central projection in the categorical coproduct.
Perhaps the spatial coproduct is some sort of coordinate-free version
of the free product mentioned in Dmitri Nikshych's answer.
(3a) For any two von Neumann algebras M and N
consider the functor F from W* to Set that sends a von Neumann algebra L
to the set of all pairs of morphisms M→L and N→L with commuting images.
The functor F preserves limits and satisfies the solution set condition, therefore it is representable.
The representing object is the categorical tensor product of M and N.
(3b) There is also the classical spatial tensor product.
I don't know any good universal property that characterizes it except
that there is a canonical map from (3a) to (3b) and its kernel corresponds
to some central projection in (3a). Perhaps there is a nice description of this central projection.
Since your monoidal structure is of the third flavor and you don't want a monoidal
structure of the first flavor, I suggest that you try a monoidal structures of the second flavor.
I suspect that the spatial coproduct of two factors is actually a factor.
You are lucky to work with factors, because in the commutative case 2=3, in particular 2a=3a and 2b=3b.
Theorem.
The map $V(A,H)\to\operatorname{Hom}(A,\mathbb{B}(H))$ is open.
We write $\omega_{\xi,\eta}$ for the linear functional $x\mapsto \langle x\xi,\eta\rangle$
and $\omega_\xi$ for $\omega_{\xi,\xi}$.
It is an elementary fact that if $\omega_\xi=\omega_{\eta}$ on a von Neumann algebra $A$,
then $a\xi\mapsto a\eta$ extends to a partial isometry $u\in A'$ such that $u\xi=\eta$.
We can trim $u^*u$ a little, if necessary, and make it satisfy $1-u^*u \sim 1-uu^*$ (Murray--von Neumann equivalence),
at the cost of $\|u\xi-\eta\|<\epsilon/2$, where $\epsilon>0$ is arbitrary small.
Then $u$ extends to a unitary element in $A'$, still denoted by $u$,
which satisfies $\|u\xi-\eta\|<\epsilon$.
The following perturbation lemma is well-known and follows from
the theory of standard form (see [Takesaki, Section IX]).
Lemma.
For any $\epsilon>0$, there is $\delta>0$ which satisfies the following.
For any von Neumann algebra $A\subset B(K)$ and any unit vectors
$\xi,\eta\in K$, if $\|(\omega_\xi-\omega_\eta)|_A\|<\delta$,
then there is a unitary element $u\in A'$ such that
$\|u\xi - \eta\|<\epsilon$.
We postpone the proof of this lemma and prove the theorem.
Let $v_0\in V(A,H)$ and an SOT neighborhood
$$G=\{ v\in V(A,H) : \forall i\ \|(v-v_0)\xi_i\|<\epsilon\}$$
be given.
Here $\xi_1,\ldots,\xi_n\in H$ are unit vectors and
$\epsilon>0$.
Take $\delta>0$ from the Lemma for $n^{-1/2}\epsilon$.
Now suppose $v\in V(A,H)$ is such that
$$\| (\omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_v
- \omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_{v_0})|_A \|<\delta/n$$
for all $i,j$.
We consider the unit vector
$\xi=n^{-1/2}\left[\begin{smallmatrix} \xi_1 & \cdots & \xi_n\end{smallmatrix}\right]^T\in H^n$ and view $\mathbb{B}(H^n)=\mathbb{M}_n\otimes\mathbb{B}(H)$.
Then
$$\|(\omega_{(1\otimes v)\xi} - \omega_{(1\otimes v_0)\xi})|_{\mathbb{M}_n\otimes A}\|<\delta.$$
Thus by Lemma, one finds a unitary element
$u\in A'\cong (\mathbb{M}_n\otimes A)'\cap\mathbb{B}(H^n)$ such that
$\|(1\otimes u)(1\otimes v)\xi - (1\otimes v_0)\xi\|<n^{-1/2}\epsilon$.
This implies that $uv\in G$, which finishes the proof.
Proof of Lemma.
We may assume $K = p(L^2A \otimes \ell_2)$, where $L^2A$
is a standard representation of $A$ and
$p\in (A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$.
Fix a unit vector $\delta_0\in\ell_2$.
There are unique vectors $|\xi|$ and $|\eta|$ in the positive cone $(L^2A)_+$
such that
$\omega_\xi=\omega_{|\xi| \otimes\delta_0}$ and
$\omega_\eta=\omega_{|\eta| \otimes\delta_0}$ on $A \otimes \mathbb{C}1$
(see [Takesaki, Theorem IX.1.2.(iv)]).
Hence $v\colon (a\otimes 1)(|\xi|\otimes\delta_0)\mapsto (a\otimes1)\xi$, $a\in A$, extends to a partial isometry in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\xi=v(|\xi|\otimes\delta_0)$ and $vv^*\le p$.
Likewise, there is a partial isometry $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\eta=w(|\eta|\otimes\delta_0)$ and $ww^*\le p$.
By the generalized Powers--Stormer inequality ([Takesaki, Theorem IX.1.2]),
one has $\| |\xi| - |\eta| \|^2 \le \|(\omega_\xi-\omega_\eta)|_A\|$.
Hence $t:=wv^*\in p(A\otimes \mathbb{C}1)'p = A'\cap \mathbb{B}(K)$ satisfies
$$\| t \xi - \eta \| = \| w(v^*\xi-w^*\eta)\|\le\| |\xi|-|\eta| \| \approx 0.$$
Let $t=u|t|$ be the polar decomposition. Since $\|t\|\le1$ and
$\|t\xi\|\approx\|\eta\|=1$, one has $|t|\xi \approx \xi$ and $u\xi\approx\eta$.
We can further replace the partial isometry $u\in A'$ with a
unitary element without affecting $u\xi$ much.
Best Answer
Since the class of approximately semifinite factors contains all injective factors and all semifinite factors and since it is closed under tensor products, it seems that this class is quite large and I would be surprised if this class could somehow be characterized.
However, there are type III$_1$ factors that are not approximately semifinite. For instance, you could take the free Araki-Woods factor associated to a measure without atoms. By Corollary C in https://doi.org/10.4171/JEMS/898 this gives you a nonamenable type III$_1$ factor $M$ with the property that for every faithful normal state $\varphi$ on $M$, the centralizer $M^\varphi$ is amenable.
It follows from this property that whenever $P \subset M$ is a semifinite von Neumann subalgebra with faithful normal conditional expectation $E : M \to P$, then $P$ must be amenable. Indeed, if $P$ is not amenable, one can choose a finite projection $p \in P$ such that $pPp$ is not amenable. Combining a faithful tracial state on $pPp$ with any faithful normal state on $(1-p)P(1-p)$, one obtains a faithful normal state $\omega$ on $P$ such that $P^\omega \supset pPp$ is nonamenable. Writing $\varphi = \omega \circ E$, we have that $P^\omega \subset M^\varphi$. So also $M^\varphi$ is nonamenable, which is a contradiction.
Since $M$ is itself nonamenable, it follows in particular that $M$ is not approximately semifinite.