I believe this cannot happen: trivially, if $\text{H}^1(M;\mathbb Z) = 0$, you clearly won't find a example of $f, g$ as desired with $I = \{2, \dots, n\}$. If $M$ is non-orientable and hence $\text{H}^n(M,\mathbb Z) = 0$, you won't find a example as desired with $I = \{2, \dots, n-1\}$.

So let as assume that $M$ is orientable and $0 \neq \text{H}^1(M)\cong\text{Hom}(\pi_1(M);\mathbb Z)$, which is torsion-free. Pick $0 \neq \alpha \in \text{H}^1(M)$ that is not a proper integral multiple of some other cohomology class. Then, by Poincaré duality, we can find $\beta \in \text{H}^{n-1}(M;\mathbb Z)$ such that $\alpha \beta = \mu_M$, the dual of the fundamental class $[M]$. Hence if $f, g \colon M \to M$ are such that $f_{\ast} = g_{\ast}$ in degrees below $n$, we also get $f_{\ast} \mu_M = (f_{\ast} \alpha)(f_{\ast} \beta) = (g_{\ast} \alpha)(g_{\ast} \beta) = g_{\ast} \mu_M$ and thus $f_{\ast} = g_{\ast}$ in all degrees; so for $I = \{1, \dots, n-1\}$ no $f,g$ as desired exist.

I believe this cannot happen: trivially, if $\text{H}^1(M;\mathbb Z) = 0$, you clearly won't find a example of $f, g$ as desired with $I = \{2, \dots, n\}$.

Suppose first that $M$ is orientable. Pick $0 \neq \alpha \in \text{H}^1(M)$ that is not a proper integral multiple of some other cohomology class. Then, by Poincaré duality, we can find $\beta \in \text{H}^{n-1}(M;\mathbb Z)$ such that $\alpha \beta = \mu_M$, the dual of the fundamental class $[M]$. Hence if $f, g \colon M \to M$ are such that $f_{\ast} = g_{\ast}$ in degrees below $n$, we also get $f_{\ast} \mu_M = (f_{\ast} \alpha)(f_{\ast} \beta) = (g_{\ast} \alpha)(g_{\ast} \beta) = g_{\ast} \mu_M$ and thus $f_{\ast} = g_{\ast}$ in all degrees; so for $I = \{1, \dots, n-1\}$ no $f,g$ as desired exist.

If $M$ is non-orientable, the same argument with $\mathbb F_2$ coefficients should work.