With m=1 dn is the sech =1/cosh
a^2 + b^2 - c^2 + 2 (-a^2 + c^2) Sech[Sqrt[-a^2 + c^2] x] == u
If your Solve[sol,u] hangs, its strange, because u is already explicit.
Solving for the inverse, you have to solve for x
In[3]:=
Solve[sol, x]
Out[3]= {{x -> ConditionalExpression[(-ArcCosh[(2 (a^2 - c^2))/( a^2 + b^2 - c^2 - u)] + 2 I \[Pi] ConditionalExpression[1, \[Placeholder]])/ Sqrt[-a^2 + c^2], ConditionalExpression[1, \[Placeholder]] \[Element]
Integers]}, {x -> ConditionalExpression[( ArcCosh[(2 (a^2 - c^2))/(a^2 + b^2 - c^2 - u)] + 2 I \[Pi] ConditionalExpression[1, \[Placeholder]])/
Sqrt[-a^2 + c^2], ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}}