Manually reversing the rotation should do the trick; there should be a formula for rotating spherical coordinate systems somewhere, but since I can't find it, here's the derivation ( ' marks the rotated coordinate system; normal geographic coordinates use plain symbols):
First convert the data in the second dataset from spherical (lon', lat') to (x',y',z') using:
x' = cos(lon')*cos(lat')
y' = sin(lon')*cos(lat')
z' = sin(lat')
Then use two rotation matrices to rotate the second coordinate system so that it coincides with the first 'normal' one. We'll be rotating the coordinate axes, so we can use the axis rotation matrices. We need to reverse the sign in the ϑ matrix to match the rotation sense used in the ECMWF definition, which seems to be different from the standard positive direction.
Since we're undoing the rotation described in the definition of the coordinate system, we first rotate by ϑ = -(90 + lat0) = -55 degrees around the y' axis (along the rotated Greenwich meridian) and then by φ = -lon0 = +15 degrees around the z axis):
x ( cos(φ), sin(φ), 0) ( cos(ϑ), 0, sin(ϑ)) (x')
y = (-sin(φ), cos(φ), 0).( 0 , 1, 0 ).(y')
z ( 0 , 0 , 1) ( -sin(ϑ), 0, cos(ϑ)) (z')
Expanded, this becomes:
x = cos(ϑ) cos(φ) x' + sin(φ) y' + sin(ϑ) cos(φ) z'
y = -cos(ϑ) sin(φ) x' + cos(φ) y' - sin(ϑ) sin(φ) z'
z = -sin(ϑ) x' + cos(ϑ) z'
Then convert back to 'normal' (lat,lon) using
lat = arcsin(z)
lon = atan2(y, x)
If you don't have atan2, you can implement it yourself by using atan(y/x) and examining the signs of x and y
Make sure that you convert all angles to radians before using the trigonometric functions, or you'll get weird results; convert back to degrees in the end if that's what you prefer...
Example (rotated sphere coordinates ==> standard geographic coordinates):
southern pole of the rotated CS is (lat0, lon0)
(-90°, *) ==> (-35°, -15°)
prime meridian of the rotated CS is the -15° meridian in geographic (rotated 55° towards north)
(0°, 0°) ==> (55°, -15°)
symmetry requires that both equators intersect at 90°/-90° in the new CS, or 75°/-105° in geographic coordinates
(0°, 90°) ==> (0°, 75°)
(0°, -90°) ==> (0°,-105°)
EDIT: Rewritten the answer thanks to very constructive comment by whuber: the matrices and the expansion are now in sync, using proper signs for the rotation parameters; added reference to the definition of the matrices; removed atan(y/x) from the answer; added examples of conversion.
EDIT 2: It is possible to derive expressions for the same result without explicit tranformation into cartesian space. The x, y, z in the result can be substituted with their corresponding expressions, and the same can be repeated for x', y' and z'. After applying some trigonometric identities, the following single-step expressions emerge:
lat = arcsin(cos(ϑ) sin(lat') - cos(lon') sin(ϑ) cos(lat'))
lon = atan2(sin(lon'), tan(lat') sin(ϑ) + cos(lon') cos(ϑ)) - φ
Your coordinates are likely in UTM (Universal Transverse Mercator) Eastings and Northings.
You can convert UTM coordinates to lat/long online using many different sites. Here in one. You can also perform this calculation in excel or inside of a database if you can work out the formula. Alternatively you can perform the conversion inside of a GIS such as QGIS or ArcGIS.
Before you can convert coordinates to lat/long you will need to know which UTM zone your coordinates fall in.
Here is an example for North America:
In this example, most of Florida falls within the UTM zone 17N.
I came across this page by searching 'batch convert UTM to lat/long'.
Best Answer
The lat/lon-Values 7486473 / 4193332 are pretty sure EPSG 3857 values which you need if you want to display your data on top of 3rd-party Background-Layers like Google or OSM.
Your other values 55.7500 / 37.6167 are in EPSG 4326. so they need to be reprojected to get their EPSG3857-Values.
You can transform them as @Joys mentioned or copy them from the map on epsg.io http://epsg.io/3857/map
To get an better idea of what needs to be transformed and what doesn't, just have a look at my jsfiddle:
http://jsfiddle.net/expedio/kLmg4uey/