**Edit:** According to the interesting comments of Michael Albanese and Nick L we revise the question as follows:


By manifold compactification of a manifold $M$ we mean a compact manifold $\tilde{M}$ which contains $M$ as an open dense subset.


Assume that $M$ is an open connected manifold which admits a manifold compactification.
Does $M$ necessarily admit a manifold compactification with zero Euler characteristic?



Compactification of open manifolds in the form of a manifold( with zero Euler characteristic)
First note that odd dimensions the question of Euler characteristic $0$ is automatic, $M$ will embed in the orientable double cover of $\tilde{M}$, which will have $\chi = 0$ by Poincare Duality.


In even dimension = $2n$ (we assume $n > 1$), we recall the following fact. If $M\_{1},M\_{2}$ are compact connected manifolds then $\chi(M\_{1} \# M\_{2}) = \chi(M\_{1}) + \chi(M\_{2}) - \chi(S^{2n}) = \chi(M\_{1}) + \chi(M\_{2}) - 2$.


To prove that some embedding into any manifold implies embedding in to Euler characteristic $0$ manifold, it is sufficient to show that any integer is equal to the Euler characteristic of some manifold of dimension $2n$, since we can do connect sums on the complement of the embedding $M \hookrightarrow \tilde{M}$ (It is easy to see that the embedding can be changed so that this complement contains an open set) to shift the Euler characteristic of $\tilde{M}$ to the correct value. 


In dimension $4$ we have $\chi(\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}) =1$ and $\chi(\mathbb{C} \mathbb{P}^{2} ) = 3$. So for any $4$-manifold $N$ connect summing with $\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}$ subtracts 1 from $\chi(N)$, connect summing with $\mathbb{C} \mathbb{P}^{2} $ adds one to $\chi(N)$, hence there is a $4$-manifold with Euler characteristic equal to any integer. In higher even dimension taking appropriate products with $\mathbb{R}\mathbb{P}^{2}$ will give the same result.


**Edit 1** As Misha points out we can now ensure the image is dense by using the fact that every connected n-manifold is a compactification of an open n-cell.


**Edit 2**. Note that the above solution holds in even dimensions atleast $4$. I will give details a counterexample in the dimension $2$ case (which was pointed out Tom Goodwillie). 


Let $S$ be an orientable surface of genus $g \geq 2$. I will show there is no embedding $I: S \setminus \{p\} \hookrightarrow S'$ (for some $p \in S$). Where $S'$ is a compact surface of Euler characteristic $0$ (i.e. a torus or a Klein bottle).


We argue by contractiction, suppose such an embedding $I$, exists. Let $C$ be the boundary of a small neighbourhood of $p$.


Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposes $S'$ as the connect sum of $S$ and compact surface $\Sigma$. In symbols: $S' = S \# \Sigma$. Hence by the formula for Euler characteristic of a connect sum given above $\chi(S') \leq -2$. 


To see this note by the classification of surfaces $\chi(\Sigma) \leq 2$ and by direct computation $\chi(S) = 2-2g \leq -2$, the fact that $\chi(S') \leq -2$ now follows directly from the formula for Euler characteristic of a connect sum. This is the desired contradiction.


