(All of the below is assuming your graphs are connected.)
Every isotopy class of graphs with $e$ edges embedded in the sphere (which is almost the plane) corresponds to a pair of permutations $\sigma,\tau$ on $2e$ letters which satisfy $(\sigma\tau)^2 = 1$, and $\sigma,\tau$ are determined up to simultaneous conjugacy (the dual graph corresponds to the pair $\tau^{-1},\sigma^{-1}$). The correspondence is explicit:
First, label each side of each edge of your graph with a number from $1$ to $2e$ in any way you please. To define $\tau$, we make a cycle for each vertex $v$, which involves only what $v$ sees as the "right hand sides" of each edge meeting $v$ (ok, fine, the clockwise sides), and rotates each one one step counterclockwise. To define $\sigma$, we make a cycle for each face $f$, which rotates all of the edge labels that $f$ meets (which face $f$) one step counterclockwise (from $f$'s point of view - so, for the infinite face, you rotate clockwise from the view inside the graph). If you've done that all correctly, then $\sigma\tau$ should be the involution that swaps the two sides of each edge. As an example, if my graph is a square, then I might get $\sigma = (1\; 2\; 3\; 4)(5\; 6\; 7\; 8)$, $\tau = (1\; 8)(2\; 7)(3\; 6)(4\; 5)$, $\sigma\tau = (1\; 5)(2\; 8)(3\; 7)(4\; 6)$.
From the permutations $\sigma,\tau$, we can reconstruct the isotopy class by gluing together the faces, edges, and vertices according to the permutations, and being pleasantly surprised when the resulting surface comes out as a sphere.
Since the way we labeled the sides of the edges was arbitrary, if we pick any permutation $s$ of the $2e$ labels, then the pair of permutations $s\sigma s^{-1},s\tau s^{-1}$ describes the same isotopy class. So if one of your graphs is described by a pair $\sigma,\tau$ and the other is described by a pair $\sigma',\tau'$, then we have to determine if there is an $s$ such that $s\sigma s^{-1} = \sigma', s\tau s^{-1} = \tau'$, and which sends the outside face to the outside face. This is significantly easier to test than it sounds: if we correctly guess the value of $s(1)$, then we can fill in all of the other values of $s$ very rapidly. If we know the value of $s(i)$, then we have $s(\sigma(i)) = \sigma'(s(i))$ and $s(\tau(i)) = \tau'(s(i))$, and since our graphs are connected, $\sigma$ and $\tau$ generate a transitive group, so this eventually determines every $s(i)$ from $s(1)$.