This is not to hard to see when we rewrite the trace in terms of the Frobenius inner product on the space of matrices $M_{m,n}(K)$ when $K\in\{\mathbb{R},\mathbb{C}\}$ defined by letting $\langle A,B\rangle=\text{Tr}(AB^*)$.
We will use that fact that in an inner product space $V$, if $a,b\in V$, then $a=b$ if and only if $\langle x,a\rangle=\langle x,b\rangle$ for all $x\in V$.
Theorem: Suppose that $\Phi(X)=A_1XB_1^*+\dots+A_rXB_r^*$.
The following are equivalent:
$\text{Tr}(\Phi(\rho))=\text{Tr}(\rho)$ for all trace 1 positive operators $\rho$.
$\text{Tr}(\Phi(X))=\text{Tr}(X)$ for all operators $X$.
$A_1^*B_1+\dots+A_r^*B_r=I$.
Proof:
$2\rightarrow 1$ is clear since we simply restrict the equation to trace 1 positive operators. $1\rightarrow 2$ follows because the vector space of trace 1 positive operators spans the vector space of all positive operators.
$2\leftrightarrow 3.$ Using the cyclic invariance of the trace, we get
$$\text{Tr}(\Phi(X))=\text{Tr}(A_1XB_1^*+\dots+A_rXB_r^*)
=\text{Tr}(XB_1^*A_1+\dots+XB_r^*A_r)=\langle X,A_1^*B_1+\dots+A_r^*B_r\rangle$$
while
$\text{Tr}(X)=\langle X,I\rangle.$
Therefore, $\text{Tr}(X)=\text{Tr}(\Phi(X))$ if and only if $I=A_1^*B_1+\dots+A_r^*B_r$.
Q.E.D.