Yes. In fact, for any diffuse abelian von Neumann algebra A (in particular A = L∞[0,1]), if M and N are factors and A ⊗ M ≅ A ⊗ N, then M ≅ N. No separability assumption on M or N is needed.

Here is a short proof.

• Let A = L∞[0,1]. Since M and N are factors, the center of A ⊗ M (resp. A ⊗ N) is A ⊗ 1. Thus any ∗-isomorphism θ: A ⊗ M → A ⊗ N maps the center onto the center and hence induces an automorphism α of A. Composing θ with the automorphism α−1 ⊗ idN of A ⊗ N, we may and do assume θ fixes A pointwise: θ(f ⊗ 1) = f ⊗ 1 for all f ∈ A.

• Choose a countable family of pairwise orthogonal nonzero projections (en)n≥1 in A with sum 1 (e.g., the characteristic functions of a partition of [0,1] into measurable subsets of positive measure). Let B = W∗(en : n ≥ 1) ⊂ A. Then B ≅ ℓ∞(N), with en the minimal central projections of B.

• Since θ fixes A pointwise, it fixes B pointwise. Consider the restriction of θ to B ⊗ M → B ⊗ N. For each n, θ fixes the central projection pn = en ⊗ 1 and hence induces an isomorphism between the corners pn (B ⊗ M) pn and pn (B ⊗ N) pn.

  But pn (B ⊗ M) pn = (en B en) ⊗ M ≅ C en ⊗ M ≅ M, and similarly pn (B ⊗ N) pn ≅ N (here we use that en is minimal in B, so en B en = C en).

• Therefore, for each n we get an isomorphism M ≅ N. In particular, M ≅ N.

This argument shows more generally that “cancellation over a diffuse abelian algebra” holds for factors: tensoring with a diffuse abelian von Neumann algebra does not lose information about the factor.