"Or related to this, is that if you had a planet which had a highly elliptical orbit- more elliptical than Mercury, but got much closer to the Sun [say 25 million miles] and went the same distance furthest from the Sun as Mercury does, it's average planetary temperature could be less than Mercury or certainly less than a orbit which was circular and was at it's average distance. Or Mercury's average temperature is less than "a Mercury" would be at Mercury average orbital distance [circular orbit]."

This is actually significant aspect for the moment- rather than traveling to other stars:)

My contention is that a planet will warm less in an elliptical orbit as compared to what planet would warm in an averaged orbital distance. And this due to spending less time at the Perihelion [or more time at Aphelion].

So, a "earth" in a highly elliptical orbital, with a Aphelion distance at 149 million km and a Perihelion at 25 million km could be "in a habitablezone".

Mercury's Perihelion distance is 46.0 million km. If Mercury had Aphelion of 149 million km [Earth's orbital distance], then it would spend less time at Perihelion than it does currently.

Now, how time does Mercury presently spend of it's 88 day year at the Perihelion portion of it orbit? And henceforth
"Perihelion portion of it orbit" is Mercury's summer.
Mercury:
Mean orbital velocity (km/s) 47.87
Max. orbital velocity (km/s) 58.98
Min. orbital velocity (km/s) 38.86
[Note: I did calculation before in some other post, and it's slightly annoying doing it again- but onward.]

Roughly, a circle 46 million km has circumference of 144.5 million km. Half of this is 72.25 million km and at it's fastest mercury traveling at 58.98 km/s.
How many seconds at 58 km/sec to go 72.25 million km-
1,245,690 seconds or 346 hrs or 14 days.
So 14 days of Summer and 74 days of winter [or dozen or so days of fall and spring and 50 days of winter].
Average Distance from Sun (10^6 km) 57.9
Shortest: 46.0
Longest: 69.8

At what point does Mercury get 57.9 10^6 km distance to the Sun? How many days before after the "noon" of summer?

In orbital distance about 158 million km from "spring" to "fall" point, so it's 79 million km from "spring" point to summer "noon".
From fall to spring point it's about 200 million km.
About 24 million km before the "spring" point, Mercury is at it's average distance from the Sun [57.9 10^6 km].
So you could say winter starts after getting further than average distance, and end when it reaches the average distance.
So this takes 48 million km out the 200 million km winter, so that makes winter having 152 million km orbital. distance.

If with a guess of an average winter orbital velocity of 40 km/sec to travel 152 million km, we get about 44 days of winter and 15 days of spring or fall and 14 days of summer

So for about 1/2 of the year [44 days], Mercury is further than the average orbit distance, for about 20 days it's around the average distance [spring/fall] and about 24 days where closer than average distance.