Would Nash equilibrium in a game when only pure strategies are allowed be also a Nash equilibrium in a game when mixed strategies are allowed as well? Would Nash equilibrium in a game when only pure strategies are allowed be also a Nash equilibrium in a game when mixed strategies are allowed as well?

Yes.
Consider a simultaneous move game. Denote the payoff function by $\pi$, the strategy of player $i$ by $s_i$, and the strategy profile of other players by $s_{-i}$. The Nash equilibrium if only pure strategies are allowed implies
$$\pi_i(s_i,s_{-i})\ge \pi_i(\hat{s}_i,s_{-i}) \forall i, \forall \hat{s}_i\neq s_i,$$
that is, $s_i$ is a best response to the profile $s_{-i}$ of all other players, and this holds for all players, by the definition of a Nash equilibrium.
Now suppose you allowed mixing (but do not add any new pure strategies to the strategy set). Consider again player $i$. Hold the strategies of all other players $-i$ fixed. Is $s_i$ - the pure strategy - still a best response in this case, and hence the above strategy profile a Nash equilibrium?
Yes, because the payoff of playing say strategy $s'$ with probability $p$ and playing $s''$ with $1-p$ is
$$p\pi_i(s',s_{-i})+(1-p)\pi_i(s'',s_{-i}),$$
i.e., it is a linear combination of the payoffs from the pure strategies. Hence,
$$\pi_i(s_i,s_{-i})\ge \pi_i(\hat{s}_i,s_{-i}) \forall i, \forall \hat{s}_i\neq s_i \\ \implies \\\pi_i(s_i,s_{-i})\ge p\pi_i(s',s_{-i})+(1-p)\pi_i(s'',s_{-i}) \forall s',s'', \forall i,$$
i.e., the payoff from playing the pure strategy (given that all other players stick with their pure strategies $s_{-i}$) is at least as large as the expected payoff from playing any mixed strategy, i.e., it remains a best response.
This is because it was a best response to begin with when only allowing pure strategies, and a mixed strategy can never do better than any pure strategy it mixes over, because the expected payoff from mixing is a linear combination of the payoffs of the pure strategies.
Final note: It may still be that additional mixed strategy equilibria exist, but the pure strategy equilibria survive in any case.
