No! Such a Cayley table will not necessarily satisfy the associative law, which is an absolutely critical property in group theory.
You can easily construct a counterexample on 5 elements.
a b c d e
b c d e a
c a e b d
d e b a c
e d a c b
Every symbol occurs exactly once in every row or column, and $a$ is the identity element, but this is not associative: $(b*b)*b = a$ while $b*(b*b) = d$. (The multiplication $xy$ is defined by looking up row $x$ and column $y$.)
In general, Cayley tables are not a good way of constructing groups. Even though Cayley tables are usually presented in beginner texts due to their resemblance to the familiar addition/multiplication tables, the associative property is difficult to check in a Cayley table. Instead, groups are usually constructed in terms of some operation which is already known to be associative, such as function composition, integer addition or multiplication, or (once you get off the ground) previously constructed groups.
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
The term Cayley table is generally restricted to finite groups. However, it’s certainly possible to generalize the idea. For a group $G$ and an element $a\in G$, the $a$ ‘row’ of the table is essentially just the function $$f_a:G\to G:b\mapsto ab\;,$$ and the $a$ ‘column’ is essentially just the function $$f^a:G\to G:b\mapsto ba\;.$$ If $G$ is countably infinite, you can visualize the Cayley table as an infinite matrix.
Let $G$ be any group, and fix $a\in G$. For each $b\in G$ you have $b=a(a^{-1}b)$, so $b$ appears in row $a$ in column $a^{-1}b$. Similarly, $b=(ba^{-1})a$, so $b$ appears in column $a$ in row $ba^{-1}$. It follows that $b$ appears in every row and column. The cardinality of the group doesn’t matter.
Added: You didn’t ask, but it’s also clear that each element of $G$ appears only once in each row and column: if $ax=ay$ or $xa=ya$, then $x=y$. Thus, each of the maps $f_a$ and $f^a$ for $a\in G$ is a bijection from $G$ onto itself, i.e., a permutation of $G$. The set of all permutations of $G$ is denoted by $\operatorname{Sym}(G)$ and is a group under composition of functions; the maps
$$G\to\operatorname{Sym}(G):a\mapsto f_a$$
and
$$G\to\operatorname{Sym}(G):a\mapsto f^a$$
are isomorphisms of $G$ to subgroups of $\operatorname{Sym}(G)$. This is Cayley’s theorem.