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 $U$.

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.