Let $\psi$ be a two-qubit state, $U \in \mathbb{C}^{2\times2}$ be a single-qubit unitary. Measure second qubit after applying U to the first qubit?

How can one show that measuring the second qubit of $$\psi$$ is the same as measuring the second qubit after applying $$U \otimes I$$ to $$\psi$$?

I know that $$\psi = \Sigma_{ij} a_{ij}|i⟩|j⟩$$, $$(U \otimes I)|\psi⟩ = \Sigma_{ij}a_{ij}U|i⟩ \otimes |j⟩$$, but I'm not sure what to do next

Let's describe the measurement that you want to perform on qubit 2 by a set of projectors $$P_i=\mathbb{I}\otimes \sigma_i,$$ where I've just written that to emphasise the structure that the measurement is on qubit 2. What is the probability, $$p_i$$, of getting result $$i$$? $$p_i=\langle\psi|P_i|\psi\rangle.$$ If, instead, you'd applied $$U\otimes\mathbb{I}$$ first, then you'd be calculating \begin{align} p_i'&=\langle\psi|U^\dagger\otimes\mathbb{I} P_iU\otimes\mathbb{I}|\psi\rangle \\ &=\langle\psi|(U^\dagger U)\otimes\sigma_i|\psi\rangle \\ &=\langle\psi|\mathbb{I}\otimes\sigma_i|\psi\rangle \\ &=\langle\psi|P_i|\psi\rangle \\ &=p_i \end{align} Hence, the probabilities are unchanged.
Another way to do this is using the partial trace. Everything that can be determined just from the second qubit is described by $$\rho_2=\text{Tr}_1|\psi\rangle\langle\psi|.$$ Basically, what you do is sum over any orthonormal basis on the first qubit, $$\rho_2=\sum_i\left(\langle i|\otimes\mathbb{I}\right)\rho\left(|i\rangle\otimes\mathbb{I}\right)$$ where I've written $$\rho$$ instead of $$|\psi\rangle\langle\psi|$$. Now, if I introduce the unitary $$U$$, we'd be calculating $$\rho_2'=\sum_i\left(\langle i|\otimes\mathbb{I}\right)U\otimes\mathbb{I}\rho U^\dagger\otimes\mathbb{I}\left(|i\rangle\otimes\mathbb{I}\right).$$ The trick here, is to change which orthonormal basis you're summing over. You can use $$|\eta_i\rangle=U|i\rangle$$ instead, so that \begin{align} \rho_2'&=\sum_i \left( \langle \eta_i| \otimes \mathbb{I} \right) (U\otimes\mathbb{I})\rho (U^\dagger\otimes\mathbb{I}) \left(|\eta_i\rangle\otimes\mathbb{I} \right). \\ &=\sum_i\left(\langle i|\otimes\mathbb{I}\right)\rho \left(|i\rangle\otimes\mathbb{I}\right). \\ &=\rho_2. \end{align} This shows the stronger result that everything that can be learnt or done to just qubit 2, without knowledge of qubit 1, is independent of the unitary applied to qubit 1.