Does $M^o=N^o$ imply that $\partial M = \partial N$?

In general, this is wrong.

Take $M=L(7,1)\times S^{2n}\times[0,1]$ and $N=L(7,2)\times S^{2n}\times[0,1]$.

Their boundaries $L(7,1)\times S^{2n}$ and $L(7,2)\times S^{2n}$ are not homeomorphic but $h$-cobordant, as proven by Milnor in

J. Milnor, Two complexes which are homeomorphic but combinatorially distinct. Ann. of Math. (2) 74 (1961) 575–590.

It follows that $\mathring{M}=L(7,1)\times S^{2n}\times \mathbb R=L(7,2)\times S^{2n}\times \mathbb R=\mathring{N}$.

However, in dimension $3$ this should be true, following from work of Edwards:

C. Edwards, Concentricity in 3-manifolds. Trans. Amer. Math. Soc. 113 (1964) 406–423.

He showed that two compact $3$-manifolds are homeomorphic if and only if their interiors are homeomorphic. A main ingerdient in his proof is (as you said) that oriented $2$-manifolds are determined by their Euler characteristic.

(Marc Kegel posted his answer just before I posted this. I will leave it because perhaps this helps elaborate some of the points)

No, this is not true in general.

Here is an example of what can happen in high dimensions. Let $N_1$ and $N_2$ be two manifolds which are h-cobordant but not homeomorphic*. We let M be the h-cobordism, which we view as a bordism from $N_1$ to $N_2$. The s-cobordism theorem (assuming $\dim M > 4$) tells us that there is an inverse bordism $M'$ from $N_2$ to $N_1$.

This is a bordism with the properties $$M \circ M' = M \cup_{N_2} M' = N_1 \times [0,1]$$ and $$M' \circ M = M' \cup_{N_1} M = N_2 \times [0,1]$$

Now we do the swindle: We consider the infinite series of composites $M \circ M' \circ M \circ M' \circ \cdots$

On the one hand we get $$(M \circ M') \circ (M \circ M') \circ \cdots = N_1 \times [0,1] \circ N_1 \times [0,1] \circ \dots = N_1 \times [0,1)$$ One the other hand we get $$M \circ (M' \circ M) \circ (M' \circ M) \circ \cdots = M \circ N_2 \times [0,1] \circ N_2 \times [0,1] \circ \cdots$$ $$ = M \circ N_2 \times [0,1) = M \setminus N_2$$

Removing the remaining $N_1$ boundary we see that $M^0 = N_1 \times (0,1)$.

So $M$ and $N_1 \times [0,1]$ have the same interior, but different boundaries.

*such manifolds are constructed, for example, in "How different can $h$-cobordant manifolds be ?" by Bjorn Jahren and Slawomir Kwasik

As the two answers already given prove, this is not true. On the other hand, one can show that these spaces cannot be distinguished by homology or homotopy groups. This greatly strengthens your statement that they have the same Euler characteristic. Considering $\pi_0$ in particular, we can say that $M$ and $N$ have the same number of connected components. Considering $H^{\ast}$ as a functor to Rings-op, you should be able to show that (for $\dim M=\dim N=2$) they have the same number of connected components of each genus.

Let $F$ be some function from topological spaces to some category closed under limits, such as $H_j$ or $\pi_j$. Let $U = M^{\circ} \cong N^{\circ}$. Consider $$\lim_{\leftarrow} F(U \setminus K)$$ where the limit is taken over all compact subsets $K$ of $U$.

Now, fix metrics on $M$ and $N$. Let $(\partial M)_{\delta}$ be an open $\delta$ neighborhood of $\partial M$ in $M$, and let $(\partial N)_{\epsilon}$ be likewise. Then $M \setminus (\partial M)_{\delta}$ and $N \setminus (\partial N)_{\epsilon}$ are each cofinal in sets of the form $U \setminus K$, so we deduce that $$\lim_{0 \leftarrow \delta} F((\partial M)_{\delta}) = \lim_{0 \leftarrow \epsilon} F((\partial N)_{\epsilon})=\lim_{\leftarrow} F(U \setminus K).$$ But, for $\delta$ small enough, $\partial M$ is a deformation retract of $(\partial M)_{\delta}$ and likewise for $N$. So, if $F$ turns deformation retracts into the identity, we deduce that $$F(\partial M) = F(\partial N).$$

When $F=H_j$ or $\pi_j$, this invariant is called the "homology at infinity" or "homotopy at infinity" of $U$.

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Here is a quicker proof that $\partial M$ is homotopy equivalent to $\partial N$ without homotopy limits. Choose $\delta_1$ small enough that, for all $\delta<\delta_1$, we have $(\partial M)_{\delta} \cong (\partial M) \times \mathbb{R}$. Choose $\epsilon_1$ small enough that the analogous condition is true, and also so that $(\partial N)_{\epsilon_1} \subset (\partial M)_{\delta_1}$. Choose $\delta_2$ and $\epsilon_2$ small enough that $(\partial N)_{\epsilon_2} \subset (\partial M)_{\delta_2} \subset (\partial N)_{\epsilon_1} \subset (\partial M)_{\delta_1}$.

Each of the inclusions above is a map in the homotopy category. Since $(\partial M)_{\delta_2}$, $(\partial M)_{\delta_1}$ and $\partial M$ are canonically homotopy equivalent, and likewise for $N$, we get maps $$\partial N \overset{\alpha_2}{\longrightarrow} \partial M \overset{\beta}{\longrightarrow} \partial N \overset{\alpha_1}{\longrightarrow} \partial M.$$ The composites $\beta \circ \alpha_1$ and $\alpha_2 \circ \beta$ are each homotopic to the identity. So $\alpha_2 \circ \beta \circ \alpha_1 = \alpha_1 = \alpha_2$ in the homotopy category, and we may denote both $\alpha_1$ and $\alpha_2$ by $\alpha$. We have shown that $\alpha$ and $\beta$ are inverse maps in the homotopy category.