Fibre transfer of $\mathbb{S}^1$-bundles

Question

Let $p:E\to X$ and $p':E'\to X'$ be two orientable $\mathbb{S}^1$-bundles. Denote their homological transfers by $p_!:H_*(X)\to H_{*+1}(E)$ resp. $p'_!$.

Now let $(u,f)$ be a bundle morphism ($u:E\to E'$ and $f:X\to X'$ and $p'\circ u = f\circ p$) and assume that $u_x:E_x\to E_x$ has degree $1$ for all $x\in X$.

Can we conclude that $u_*\circ p_!=p'_!\circ f_*$ as maps in homology?

Edit: I think I solved it, and it immediately generalises to $\mathbb{S}^k$-bundles:

1. Let $q:(D,E)\to X$ be the associated $(\mathbb{D}^2,\mathbb{S}^1)$-bundle, $v:(D,E)\to (D',E')$ the extended morphism and $\tau\in H^2(D,E)$ the Thom class. Our condition tells us that $\tau=v^*\tau'$.
2. Consider the connecting homomorphism $\delta_*:H_*(D,E)\to H_{*-1}(E)$ resp. $\delta_*'$. By naturality $u_*\circ \delta_* = \delta'_*\circ v_*$.
3. We have the Thom isomorphism $t:H_*(D,E)\to H_{*-2}(X),x\mapsto q_*(\tau\frown x)$ and $t'$. We see that $(t')^{-1}\circ f_* = v_*\circ t^{-1}$ by $$(f_*\circ t)(x) = f_*q_*(\tau\frown x) = q'_*v_*(v^*\tau'\frown x)=q_*(\tau'\frown v_*x)=(t'\circ v_*)(x)$$
4. In the Gysin sequence, $p_!$ appears as $p_!=\delta^*\circ t^{-1}$. Thus, finally $$u_*\circ p_! = u_*\circ \delta_*\circ t^{-1} = \delta'_*\circ v_*\circ t^{-1} = \delta'_*\circ (t')^{-1}\circ f_* = p'_!\circ f_*.$$