TL;DR: about 5 times current eccentricity for a deviation by 5 degrees from the mean... but the devil is in the celestial mechanics and climate model details.
It would not quite work, because the length of seasons would be uneven. Currently Earth's north and south hemispheres get nearly exactly as much or little sunlight as the other. But on the eccentric Earth the hot "summer" period would be short while the cool "winter" period would be long since planets spend more time in the outer part of an elliptic orbit.
There is another complication: absorbed sunlight heats up air, water and soil over time causing a lag: the warmest part of summer and the coldest part of winter comes after the time of the highest solar influx.
To add to the modelling woes, the actual temperature of the planet is strongly affected by the greenhouse effect: while CO2 is not going to do anything weird, as the entire planet cools water vapour (a potent greenhouse gas) will decline as rain and snow, and the albedo (how much light is reflected by clouds and ice) will increase.
That said, here is a simplistic model ignoring the atmosphere inertia:
The distance to the star varies as $$r(\theta)=\frac{a(1-e^2)}{1+e\cos \theta}$$ where $a$ is the semimajor axis, $e$ the eccentricity and $\theta$ the "true anomaly" (annual winner for worst name of parameter since 1609). To convert it to and from time you need to solve Kepler's equation numerically.
The temperature in a zero-dimensional atmospheric model is $$T=\left( \frac{ I_0(1-a)}{4 \sigma \epsilon}\right )^{1/4}$$ where $I_0$ is the solar constant at the top of the atmosphere, $a=0.3$ the albedo, $\epsilon=0.78$ the effective emissivity in IR. If we let $I_0(\theta)=I_0/r(\theta)^2$ (measuring orbits in AU) we get a temperature dependency scaling like $T(\theta)=T_0/\sqrt{r(\theta)}$.
So, at perihelion the temperature is $$T(0)=T_0\sqrt{\frac{1+e}{1-e^2}}$$ and the aphelion temperature $$T(\pi)=T_0\sqrt{\frac{1-e}{1-e^2}}.$$ So given a desired $T_{high}$ and $T_{low}$ one can now solve for $e$ to get $$e=\frac{(T_{high}/T_{low})^2-1}{(T_{high}/T_{low})^2}.$$
If we want a 5 degree difference from a mean $T_0=288$ K this means $e\approx 0.067$, about 5 times the current eccentricity. This is probably small enough that the time asymmetry issues will be small enough to ignore (they matter more for more eccentric orbits).
