Yep, you're right
if a group G of order 4 has an element of order 4, say x, then {x,x^{2},x^{3},x^{4} = e} must be all of G, so G = <x>, and is isomorphic to C_{4}.
if not, then all 3 non-identity elements must be of order 2 (the only other divisor (besides 1) of 4 left). pick any two of them, say a and b.
now <a> and <b> are two subgroups of order 2. it is straightforward to show that G is isomorphic to <a> x <b> (which is obviously isomorphic to C_{2} x C_{2}):
e --> (e,e)
a --> (a,e)
b --> (e,b)
ab --> (a,b) is the isomorphism.
(proving ba = ab is the only "non-obvious thing" (which must be true if we have an isomorphism since (a,e)*(e,b) = (ae,eb) = (a,b) = (ea,be) = (e,b)*(a,e)). it is perhaps easiest to show that ba = e,a, or b leads to a contradiction).