I want to find an algebra isomorphism $$f : \mathbb{C}\otimes_\mathbb{R}\mathbb{C}\rightarrow \mathbb{C}\oplus \mathbb{C}. $$
Define $$a_1=1\otimes 1,\ a_2=1\otimes i,\ a_3=i\otimes 1,\ a_4=i\otimes i $$
That is, $a_2a_3=a_4$. Now, $(1,-1)= f(1\otimes (-1))=f((-1)\otimes 1) = (-1,1 ). $ So this is not a suitable map. Thanks in anticipation.
The idea of the proof is that $\Bbb C/\Bbb R$ is a finite separable field extension of degree two, so that $\Bbb C\otimes_\Bbb R \Bbb C$ splits as two copies of $\Bbb C$. This happens in general, if $L/k$ is finite separable of degree $n$ you should be able to show that $L\otimes_k L$ is isomorphic to $n$ copies of $L$ for $L$ algebraically closed.
Consider the element $e=\frac 1 2(1\otimes 1-i\otimes i)$. Then under the multiplication map $\mu:\Bbb C\otimes \Bbb C\to\Bbb C$ $\mu(e)=1$. Check that $e$ annihilates the kernel of $\mu$, which proves $\Bbb C$ is a separable $\Bbb R$-algebra. This implies that the tensor product is semisimple so it must be a product of copies of $\Bbb C$, and since it is $4$ dimensional over $\Bbb R$, it must be $\Bbb C\times\Bbb C$.
As an $\Bbb R$ vector space, $\Bbb C$ is generated by $1$ and $i$, so that the kernel of $\mu$ is generated by $i\otimes 1-1\otimes i$, and $$e(1\otimes i-i\otimes 1)=1\otimes i+i\otimes 1-i\otimes 1-1\otimes i=0$$
A concrete isomorphism is given by mapping $e\mapsto (1,0)$ and $f=1-e\mapsto (0,1)$. Note that $f^2=f,e^2=e,ef=fe=0$ and $e+f=1$.