Your hypothesis (modifying the topology to produce a non-Hausdorff topological manifold from a Hausdorff one) is impossible, hence the answer to the question you asked is vacuously "yes". 

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Let $(M,\tau)$ be your topological manifold, and denote by $\tau'$ the coarser topology, which you asserted is locally Euclidean. 

First we remark that the identity map, regarded as a function from the topological space $(M,\tau)\to (M,\tau')$ is continuous; this is a property of $\tau' \subseteq \tau$ (your coarser assumption). 

Given $p\in M$, by the locally Euclidean assumption we may choose an open (w.r.t $\tau'$) neighborhood $O'$ that is homeomorphic to $E'$, some Euclidean space. Note that $O'$ is also open w.r.t. $\tau$. 

Similarly there exists another open (w.r.t $\tau$) neighborhood $O$ that is  homeomorphic to $E$, another Euclidean space. 

Now $O' \cap O$ is an intersection of open sets in the $\tau$ topology, and so is a relatively open subset of $O$, and hence is homeomorphic to some open set $\bar{E}\subseteq E$ in an Euclidean space. 

Consider the chain of mappings (for convenience I specify the topology for the middle two spaces)

$$ \bar{E} \to (O'\cap O,\tau) \to (O',\tau') \to E' $$

with the first map the homeomorphism of $O$ and $E$ and the final map the homeomorphism of $O'$ and $E'$. The middle map is the identity map on $M$. 

This chain gives an injective continuous map from $\bar{E}$ to $E'$, hence by [invariance of domain](https://en.wikipedia.org/wiki/Invariance_of_domain) it is also a homeomorphism to its image. 

In particular, this shows the following:

> Given any $p\in M$, there exists an open neighborhood (w.r.t $\tau$) $\Omega$ of $p$ such that every open subset $U\subseteq \Omega$ (w.r.t $\tau$) is also open w.r.t. $\tau'$. 

Now, assume that $\tau$ is Hausdorff. Given $p\neq q\in M$ let $\Omega_p$ and $\Omega_q$ be as in the previous result. Let $U_p$ and $U_q$ be disjoint open sets (w.r.t. $\tau$) that separate $p$ and $q$. Then $U_p\cap \Omega_p$ and $U_q\cap \Omega_q$ are still disjoint, separates $p$ and $q$, and are both open in the $\tau'$ topology. This shows that $\tau'$ must also be Hausdorff.