How to prove that this map is well-defined, surjective and 1-1:
$$1\otimes a+i\otimes b+j\otimes c+ij\otimes d \mapsto a \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} + b \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix} + c \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix}$$
Where $a,b,c,d$ are complex numbers and this map gives the isomorphism $\Bbb{H\otimes_R C }\to M_2(\mathbb C)$
Another question: Why a map that maps a basis to a basis must be an isomorphism?is there a proof or a theorem that say this? Should I use that theorem here?
First, you ask: "Why a map that maps a basis to a basis must be an isomorphism?" Well, remember these are vector spaces over $\mathbb C$ in particular. So the correct statement is that a linear map that maps a basis to a basis is a (linear) isomorphism. This does not by itself guarantee that they are isomorphic as rings too. Okay, on to the problem.
Well-defined
$\mathbb H$ is an $\mathbb R$-algebra with basis $\{1, i, j, ij\}$. We then know by "extension of scalars" that $\mathbb H\otimes_{\mathbb R} \mathbb C$ is a $\mathbb C$-algebra with basis $\{1\otimes 1, i\otimes 1, j\otimes 1, ij\otimes 1\}$. But that's the same as saying that an $x\in \mathbb H\otimes_{\mathbb R} \mathbb C$ can be written uniquely as $$ x = 1\otimes a+i\otimes b+ j\otimes c+ ij\otimes d $$ with $a,b,c,d\in \mathbb C$, so your map $\varphi:\mathbb H\otimes_{\mathbb R} \mathbb C \to M_2(\mathbb C)$ is well-defined.
Homomorphism of algebras
It is easy to see that $\varphi$ is linear ($a,b,c,d$ are scalars). What remains is to show $\varphi(1)=1$ (which is evident) and $\varphi(xy)=\varphi(x)\varphi(y)$. Using linearity of $\varphi$, we only need to check this on base vectors. For example: $$ \varphi(i\otimes 1)\varphi(j\otimes1) = \begin{pmatrix} i&0\\0&-i\end{pmatrix} \begin{pmatrix} 0&1\\-1&0\end{pmatrix} = \begin{pmatrix} 0&i\\i&0\end{pmatrix} = \varphi(ij\otimes 1) $$ and so on. (This amounts to showing that the matrices generate the quaternion group.) Alternatively, we could take the product of two general elements and do it all in one go without appealing to linearity.
Bijective
Now we just need to show $\varphi$ is bijective. Since $\varphi$ is a linear map, we only need to show that the four matrices are a basis of $M_2(\mathbb C)$. It's obvious that another basis for $M_2(\mathbb C)$ is $$ \mathscr B = \left\{ \begin{pmatrix} 1&0\\0&0\end{pmatrix}, \begin{pmatrix} 0&1\\0&0\end{pmatrix}, \begin{pmatrix} 0&0\\1&0\end{pmatrix}, \begin{pmatrix} 0&0\\0&1\end{pmatrix} \right\} $$ In particular $M_2(\mathbb C)$ is $4$-dimensional over $\mathbb C$. And now we can prove linear indepence by writing our matrices as (column) vectors over $\mathscr B$ and take the determinat: $$ \det \begin{pmatrix} 1&i&0&0\\0&0&1&i\\0&0&-1&i\\1&-i&0&0 \end{pmatrix} = 4 \ne 0 $$ Done! We conclude that $\varphi$ maps a basis to a basis, so it is bijective and thus an isomorphism.