Let $K$ be a field of characteristic zero. It's true that if $R$ is a $K$-algebra then $R\otimes_{\mathbb{K}}\mathbb{K}[[t]]\cong R[[t]]$ with the natural inclusion inducted by $x\otimes t^j$$\mapsto$$xt^{j}$?
It's clearly injective because if $\sum_{j}r_{j}\otimes c_{j}t^{j}\mapsto 0$ then for every j $r_{j}c_{j}=0$ and so for every j $r_{j}=0$ or $c_{j}=0$
But I'm not sure about the surjectivity.
It is in general not surjective.
More precisely, the map $\phi: R\otimes K[[t]] \rightarrow R[[t]]$ sending $x \otimes t^j$ to $xt^j$ is surjective if and only if $R$ is finite dimensional as $K$-vector space.
In one direction, if $R$ is finite dimensional, then you can pick a basis of $R$ and decompose every power series in $R[[t]]$ according to this basis.
In the other direction, assume that $R$ is not finite dimensional. Note that if $\sum r_it^i$ is any power series in the image of $\phi$, then it can be written as a finite sum $\sum x_j f_j$ with $x_j \in R$ and $f_j \in K[[t]]$ and hence the sub-$K$-vector space of $R$ generated by all $r_i$ is contained in the sub-$K$-vector space generated by all $x_j$, which is finite dimensional.
But since $R$ is infinite dimensional, we can pick a sequence of liearly independent elements $(r_i)_i$ and form a power series $\sum r_i t^i$, which is then not in the image of $\phi$.
This uses nothing about the ring structure of $K[[t]]$ or $R[[t]]$. The proof works the same with $K[[t]]$ and $R[[t]]$ replaced by the infinite product $K^\Bbb N$ and $R^\Bbb N$, respectively.