Let $P:=\pmatrix{a&c\\b&d}$ with tr($P$)=$a+d.$ Let us consider operator $T$ defined by:
$$\tag{0}T(A)=PA$$
First solution:
Setting $A:=\pmatrix{x&z\\y&t}$ and $PA:=\pmatrix{x'&z'\\y'&t'}$, formula (0) is equivalent to:
$$\begin{cases}x'&=&ax+cy\\y'&=&bx+dy\\z'&=&az+ct\\t'&=&bz+dt\end{cases}$$
Otherwise said, to:
$$\tag{*}\pmatrix{x'\\y'\\z'\\t'}=\pmatrix{a&c&0&0\\b&d&0&0\\0&0&a&c\\0&0&b&d}\pmatrix{x\\y\\z\\t}$$
The trace of this matrix is clearly twice the trace of $P$.
Remark: The trace of operator $T$ has been computed with a certain matrix, but it is the same with respect to any other since it is an invariant.
Second solution (that has strong connections with the previous solution):
Consider the canonical basis on the vector space of $2 \times 2$ matrices $\frak{M_2}$ with entries in field $F$:
$$E_1=\pmatrix{1&0\\0&0}, \ E_2=\pmatrix{0&0\\1&0}, \ E_3=\pmatrix{0&1\\0&0}, \ E_4=\pmatrix{0&0\\0&1}.$$
Let us compute the images of the $E_k$s by operator T:
$$T(E_1)=\pmatrix{a&c\\b&d}\pmatrix{1&0\\0&0}=\pmatrix{a&0\\b&0}=aE_1+bE_3.$$
In a similar way:
$$T(E_2)=\pmatrix{c&0\\d&0}=cE_1+dE_2$$
$$T(E_3)=\pmatrix{0&a\\0&b}=aE_3+bE_4$$
$$T(E_4)=\pmatrix{0&c\\0&d}=cE_3+dE_4.$$
Thus the matrix of operator $T$ with respect to the canonical basis is:
$$\tag{1}[T]=\pmatrix{a&c&0&0\\b&d&0&0\\0&0&a&c\\0&0&b&d}$$
as obtained in the first solution, with the same conclusion.
Third (shorter) solution: The RHS of (0) can be transformed under the following form, using concepts and results on Kronecker product that can be found in this reference:
$$\tag{2}PA=PAI_2=(I_2^T \otimes P)vec(A)=(I_2 \otimes P)vec(A)$$
where vec$(A)$ is obtained by piling up the two colums of $A$.
Formula (2) clearly amounts to formula (*).
An important thing is that, with this method, we have not to give an explicit form to the matrix of transformation $T$; it suffices to use the general formula: tr$(A \otimes B) =$ tr$(A) $tr$(B)$ (see the same Wikipedia reference).
Were the same question been asked for $n \times n$ matrices, we would have at once the answer $n \ $tr$(P)$.
Your proof is absolutely fine. Here go my alternative approach (which is essentially the same):
Denote the zero column vectors, of both $V$ and $W$, by $\textbf0$. For $j=1,\dots,n$, let $e_j$ be the $n\times1$ matrix which has a $1$ in the $j$-th place and $0$ elsewhere. Then, for all $j$,
$$T(e_j) = Ae_j = A_j.$$
So, if $T$ is the zero transformation, the left hand side of the above equation is $\textbf0$, meaning that, each $A_j$ is $\textbf0$.
And, if $A$ is the zero matrix, the right hand side of the above equation is $\textbf0$, meaning that $T$ maps the vectors $e_1,\dots,e_n$ into $\textbf0$, which only happens if $T$ is the zero transformation.
Best Answer
Your argument for the $[\Longleftarrow]$ part is correct. This is because $0\cdot a=0$ for all $a\in F$, $0+0+\cdots+0=0$ and $AX$ has entries consisting of additions and multiplications of these types when $A=0$.
For $[\Longrightarrow]$, consider taking $X$ as the vectors $e_i$, where $e_i$ is the vector consisting of $1$ as its $i$-th entry and $0$ for its other entries. Taking $X=e_i$ will show that the $i$-th column of $A$ is $0$. Hence $A=0$.