Let $L/K$ be (any) field extension. Is it true that we have an isomorphism $K[[t_1, \dots, t_n]] \otimes L \cong L[[t_1, \dots, t_n]]$?
We know that if we remove the field assumption, this is not true. Take, for example, $\mathbb{Z} \subset \mathbb{Q}$.
On the other hand, if the extension is a finite field extension this is true since then $L$ would be a free $K$ module of finite rank. (If I am not mistaken, by taking the direct limit over all finite extension, we can also show that this is true if $L/K$ is algebraic).
For the canonical map: $$A=K[[t_1,\ldots,t_n]]\otimes_K L$$ is the subring of $B=L[[t_1,\ldots,t_n]]$ of formal power series whose coefficients lie in a finite dimensional $K$-vector subspace of $L$.
So $A$ is the whole of $B$ iff $L/K$ is a finite extension.
For a possible non-canonical isomorphism:
Both $A$ and $B$ are local rings (their non-units form a maximal ideal) and $B$ is complete $\cong\varprojlim_{k\to \infty} B/\mathfrak{m}_B^k$.
But $A$ is non-complete if $L/K$ is not a finite extension (if $\ell_j$ is a $K$-linearly independent sequence of elements of $L$ then $\sum_{j\ge 0} t_1^j\otimes \ell_j $ is in the completion $\varprojlim_{k\to \infty} A/\mathfrak{m}_A^k$ but not in $A$)
So $A$ and $B$ are not isomorphic as abstract rings.