Mathematicians typically don't use the word "homomorphic" that way, for the reason (as pointed out above) that for any two groups, $G,H$, the $0$-map (or trivial map):
$0:G \to H$ given by $0(g) = e_H$ for all $g \in G$ (map everything in $G$ to the group identity of $H$), always exists.
In any case, the existence, or non-existence, of a (non-trivial) homomorphism between two groups is a poor yardstick of similarity; for distinct primes $p, q$ there is no non-trivial group homomorphism $f: \Bbb Z/p\Bbb Z \to \Bbb Z/q\Bbb Z$, but very few mathmeticans would claim this groups are "dissimilar" (they are both simple cyclic groups).
It is (perhaps) more fruitful to think of a homomorphism $f: G \to f(G) \subseteq H$ as "preserving" partial information about $G$ "inside" of $H$. Some information about $G$ can be lost, as homomorphisms are not always one-to-one (when they are one-to-one, this is a highly desirable situation, as we have a "copy" of $G$ inside $H$, and $H$ may be "easier" to understand than $G$).
Groups can be very different from one another (as opposed to, say, finite-dimensional vector spaces over the reals, which are all rather alike). This is mostly because groups have so few rules, that some wildly disparate sets and operations can still qualify to be groups. Mathematicians have developed classifications, such as order, simplicity, etc. to help distinguish groups by type, and it turns out this is....complicated.
It should be clear that one (and only one) of the elements is the identity element of your respective tables, and that an isomorphism must map these elements to each other. So the question remains, what can we deduce from the remaining three elements?
You are correct that IF the two groups are isomorphic, that the mapping between them must be a bijection on $\{a,b,c\}$. Fortunately for you, there are only six such mappings:
The identity map (if they are not the same table, we can rule this out),
Three transpositions: swap $(a,b)$, swap $(a,c)$ or swap $(b,c)$
Two "cyclical" maps: $a \to b \to c \to a\dots$ ,
$a \to c \to b \to a\dots$
But there is a faster way: deduce how many elements $x$ have the property that:
$x\ast x = e$ (this amounts to finding out how many times $e$ (or its analogue) occurs on the diagonal of the table).
This is the same as counting the number of elements of order $2$.
While having the same number of elements of order $2$ does not guarantee two groups are isomorphic; what is true, is that having differing numbers of elements of order $2$ proves two groups are not isomorphic. This is something worth remembering, as it often proves to be a useful short-cut.
And in the special case of groups of order $4$, it indeed settles the matter (because $4$ has a limited set of divisors, being a prime power).
If your two tables have the same number of elements of order $2$, and they are indeed group tables (it turns out showing associativity is "hard" from just the table information), they will be isomorphic. But I caution you here, the isomorphism between them will not be unique, in fact you should be able to find at least two such permutations (bijections) of $\{a,b,c\}$ that will work.
This underscores one of the inherent difficulties with Cayley tables: it is possible to have several distinct tables that represent "the same" (that is, isomorphic) group. One kind of group of order $4$ can have six tables, the other kind can have two. So in this sense Cayley tables are "inefficient conductors of information".
If we list the orders of the respective elements, there are only two lists (when ordered in increasing order):
$\{1,2,2,2\}$
$\{1,2,4,4\}$
It turns out for groups of order $4$ (but the proof of this is too long for this post) that this is all there is, and that any two groups of order $4$ of the same "order type" are isomorphic. I would recommend you try to prove this for yourself, it's interesting.
If the two groups are of the first order type, any permutation of $(a,b,c)$ yields an isomorphism. If the two groups are of the second type, there are only two possibilities, the identity, or a swap of the two elements of order $4$ (assuming you use the same "alphabet" for your two tables).
Best Answer
Here is a trivial counter-example.
Let $G$ and $H$ be two groups of the same order, but not isomorphic. For any pair of groups, there is always the $0$ homomorphism between them, i.e. the map $0: G \to H: G \ni g \mapsto e \in H$. This is a homomorphism trivially, and so there is a homomorphism from $G$ to $H$, but somehow it is not a useful homomorphism. It's not even close to an isomorphism.
I expect the relationship you're after is given by the First Isomorphism Theorem, which in a sense tells you how a homomorphism can be 'close' to an isommorphism. In particular, if $\phi: G \to H$ is a homomorphism, this theorem asserts that $G/\ker \phi \simeq H$.
Edit: If by homomorphic you mean that $G$ has a homomorphic image which is ONTO $H$, then yes your claim is true, since for groups with the same finite order, being surjective implies being injective.