Hint: Both sides are $\mathbb R$-linear in $(a,b,c,d)$, so you have to check multiplicativity only on the $\mathbb R$-basis $\{1,i,j,k\}$ of $\mathbb H$, that is one only has to check whether
$$ I := \def\p#1#2#3#4{\begin{pmatrix}#1\\#3\end{pmatrix}}\p i00{-i}, J := \p 01{-1}0, K := \p 0ii0 $$
fulfill $I^2 = J^2 = K^2 = IJK = -\mathrm{Id}$.
Given a quaternion $q$ and a complex number $\lambda$, for scalar multiplication you can either apply $\lambda$ to $q$ from the left (i.e. $\lambda q$) or from the right (i.e. $q\lambda$). This means you can interpret $\mathbb{H}$ as a left complex vector space or as a right complex vector space. Now consider the left and right multiplication maps
$$ L_p(x)=px, \qquad R_p(x)=xp. \tag{1}$$
Then $L_p(x\lambda)=L_p(x)\lambda$ for all complex numbers, so $L_p$ is a linear transformation of $\mathbb{H}$ as a right complex vector space. And $R_p(\lambda x)=\lambda R_p(x)$, so $R_p$ is a linear transformation of $\mathbb{H}$ as a left complex vector space.
However, $L_p$ is not linear if we treat $\mathbb{H}$ as a left complex vector space, and $R_p$ is not linear if we treat $\mathbb{H}$ as a right vector space. This is because $\mathbb{H}$ is not commutative. (Exercise.)
Using $\{1,\mathbf{j}\}$ as a basis for $\mathbb{H}$ as a complex vector space, you would write an arbitrary quaternion as $z+w\mathbf{j}$ if you're thinking left vector space, and write $z+\mathbf{j}w$ if you're thinking right vector space. (Also notice $w$ comes before $z$ in the alphabet, so we're backwards alphabetically.) You're thinking left vector space but you're examining the right-linear transformation $L_p$. That's problematic.
If you look at $L_p$ and think right vector space, you should get
$$ \begin{bmatrix} z & -\overline{w} \\ w & \phantom{-}\overline{z} \end{bmatrix}. \tag{2} $$
(Exercise.)
Representing quaternions as matrices using $R_p$ is problematic since $R_p\circ R_q\ne R_{pq}$. Indeed, this would be a function $\mathbb{H}\to M_2(\mathbb{C})$ which is not a homomorphism but rather an anti-homomorphism, i.e. it satisfies $R_p\circ R_q=R_{qp}$ (the order of multiplication is reversed). In order to turn it into a homomorphism proper, one must either (pre)compose it with an anti-automorphism of $\mathbb{H}$ or (post)compose with an anti-automorphism of $M_2(\mathbb{C})$. Quaternion conjugation satisfies $\overline{pq}=\overline{q}\,\overline{p}$ and matrix transpose satisfies $(AB)^T=B^TA^T$, so we can use these as anti-automorphisms.
In the case of quaternion conjugation, we can take $p=z+w\mathbf{j}$, then its conjugate $\overline{p}=\overline{z}-w\mathbf{j}$, then the matrix of $R_{\overline{p}}$ you can calculate (exercise) to be
$$ \begin{bmatrix} \overline{z} & \overline{w} \\ -w & z \end{bmatrix}, \tag{3} $$
and if instead you just calculated $R_p$'s matrix and took the transpose you'd get $(2)$ again.
Complex conjugation (applied entry-wise to a matrix) is an automorphism of $M_2(\mathbb{C})$, i.e. it satisfies $\overline{AB}=\overline{A}\,\overline{B}$, so we may compose it with any of the representations above to get another valid representation.
Best Answer
Personally I sympathize with your doubt about the definition of a quaternion. I would have preferred a more explicit definition of the term quaternion by itself (as opposed to the compound term quaternion algebra). We cannot always assume that mathematical terms combine or uncombine in intuitively obvious ways. Possibly your internet source (where you got the solution) has a better definition if you look for it.
Take a unit quaternion $h$ and add it to itself: $h + h.$ The result is not a unit quaternion, as you can confirm by finding its determinant. So unit quaternions are not closed under addition, and cannot be a group under addition.
Associativity is a very basic property of matrix multiplication. See https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk05/05_associativity_of_matrix_multiplication.html for example.