Let $e$ be the orthogonal projection onto $\ker(x)$. If the result is not true, then there is $\xi\otimes\eta \in H_1\otimes K$ with $$ (e\otimes 1) v_1 (e\xi\otimes\eta) \ne v_1 (e\xi\otimes\eta), $$ because the linear span of such vectors in dense in $H_1\otimes K$. Here $K$ is some auxiliary Hilbert space such that we can regard $A\subseteq\mathcal B(K)$. Similarly, there is $\xi'\otimes\eta'\in H_1\otimes K$ with $$ \bigl((e\otimes 1) v_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr) \not= \bigl( v_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr). $$ I write $(\cdot\mid\cdot)$ for the inner-product.
The map $T:\mathbb C \rightarrow K$; $\alpha\mapsto\alpha\eta'$ is bounded, and so there is an adjoint $T^*:K\rightarrow\mathbb C$ which is simply $K \ni \zeta \mapsto (\zeta\mid\eta')$. Then $1\otimes T^*:H_1\otimes K \rightarrow H_1\otimes\mathbb C \cong H_1$ is bounded. Consider $$ \xi'' = (1\otimes T^*)v_1(e\xi\otimes\eta) \in H_1. $$ For any $\xi_0\in H_1$ we have that $(\xi''\mid\xi_0) = (v_1(e\xi\otimes\eta)\mid\xi_0\otimes\eta')$ and hence in particular $$ (\xi''\mid\xi') \ne (\xi''\mid e\xi'). $$
Notice that $(x\otimes 1)v_1(e\otimes 1) = v_2(xe\otimes 1)=0$ and so $(x\otimes 1)v_1(e\xi\otimes\eta) = 0$. Hence $$ 0 = \bigl((x\otimes 1)v_1(e\xi\otimes\eta)\bigm|\xi_0\otimes\eta'\bigr) = \bigl(v_1(e\xi\otimes\eta)\bigm|x^*\xi_0\otimes\eta'\bigr) = (\xi''\mid x^*\xi_0) = (x\xi''\mid\xi_0), $$ for any $\xi_0\in H_1$. This shows that $x\xi''=0$ so $\xi''\in\ker(x)$ so $e\xi''=\xi''$, which gives the required contradiction.
