Here's my understanding of the Nimbers and of your question.  Actually, you'll see that I do need to beg the question somewhere, but since you do know a proof of the bitxor rule, perhaps that's allowed --- then my answer can be understood as a proof that your proof implies your proof is natural.  (Said that way it sounds like an application of [Lob's theorem](http://www.yudkowsky.net/assets/44/LobsTheorem.pdf), or perhaps a converse....)

  By "Nimbers" I mean Nim games with Conway's game addition (put two games next two each other; on your turn, you choose one of the two boards to play on) modulo the second-player-win Games.  By definition, a "game" is one where you lose when you cannot make a turn.  The Nimbers are the classes of single-column Nim games.

The zeroth observation is that addition (henceforth "$+$") is commutative, and that for any game $g$, the game $-g$ in which the roles are reversed is its inverse.

The first observation, then, is that impartial games, and multi-column Nim games in particular, are 2-torsion: for any Nim game $g$, we have $g + g = g$.  Thus the group generated by the Nimbers is a vector space over $\mathbb F_2$.

The next ingredient I don't really have an a priori reason for, which is that the sum of any two Nimbers is a Nimber. Actually, proving this is probably just about the same as finding the bitxor formula, so perhaps my whole story is question-begging.  But let's assume that this second ingredient is just an "observation".

The third observation is the following.  Let $G_k$ denote the group generated by the Nimbers $1,\dots,k$.  If you allow the second observation, then it is not hard to see that if $n \in G_k$, then for every $m < n$, $m\in G_k$.  Indeed, if $n\in G_k$, then I can write $n = \sum a_i$ for some sum of Nimbers with $a_i \leq k$.  Let's play the game $n + \sum a_i$, which is a second-player win by assumption.  Being magnanimous, I'll go first.  On my turn I turn $n$ into $m$.  Now you definitely have a move the return the sum to $0$.  It definitely doesn't involve the pile I touched, so it must involve dropping one of the $a_i$s to an $a_i' < a_i$.  But $a_i$ was one of our generators in $1,\dots,k$, and so $a_i'$ is also one of those generators.

Now we can put the observations together to describe the structure of the Nimbers.  We have $G_0 = \{0\}$ and $G_1 = \{0,1\}$ is the group of order $2$.  By induction, the set of Nimbers $G = \{0,1,\dots,2^k-1\}$ is closed under Nimber addition.  Consider $G_{2^k}$.  It is an $\mathbb F_2$-vector space generated by $G$, which has $2^k$ elements, and by one more element.  Thus $|G_{2^k}| = 2^{k+1}$.  Thus $G_{2^k} = \{0,\dots,2^{k+1}-1\}$.  The induction can then continue.

So the Nimbers are naturally organized as an $\mathbb N$-filtered $\mathbb F_2$-vector space: $$\{0\} \subset \{0,1\} \subset \{0,1,2,3\} \subset \{0,1,2,3,4,5,6,7\} \subset \dots \subset \{0,\dots,2^{k-1}\} \subset \dots.$$
This doesn't completely pin down the addition, but it makes bitxor seem very likely.  For example, it implies that if $m,n \in \{2^{k-1},\dots,2^k-1\}$, so that they have the same leading digit mod $2$, then their sum $m+n < 2^{k-1}$, and on the other hand if $m < 2^{k-1}$ and $n \in \{2^{k-1},\dots,2^k-1\}$, then $m+n \in \{2^{k-1},\dots,2^k-1\}$.  This gives the bitxor rule in the leading digit.

Of course, this analysis still allows lots of group structures on $\{0,\dots,2^k-1\}$.  The rule is only that the structure has to extend the one on $\{0,\dots,2^{k-1}-1\}$.  You can write down ad hoc group structures by twisting the given one by any permutation of $\{2^{k-1},\dots,2^k-1\}$.  To completely pin down the bitxor group law requires playing a bit more with the third observation, I think.  Let's see if we can do it.  We know that the bitxor rule applies to Nimbers $N < 2^k$, by induction.  We also know that $2^k + 2^k = 0$, so it applies to $N \leq 2^k$.  To prove the claim, it suffices to prove that the Nimber of height $2^k+2^j$ for $j<k$ is equal to the Nim addition $2^k + 2^j$.  Everything else will follow from linear algebra.  But now all I need to do is to tell you a second-player-win strategy for the three-column Nim game with heights $2^j$, $2^k$, and $2^k+2^j$.  Well, let's suppose you play on column $2^j$.  Then I have such a strategy by induction in $j$ (and linear algebra).  Suppose you play on column $2^k$.  Then you make it into something in $\{0,\dots,2^k-1\}$, and adding $2^j$ keeps me in that group, so I just need to drop the column of height $2^k+2^j$ to match.  Finally, suppose you play on column $2^k+2^j$.  If you leave it above $2^k$, I'll play on column $2^j$ to match, and use induction in $j$ to know that that works.  If you take it below height $2^k$, I'll drop the $2^k$-column to make the sum come out, using bitxor in the group $\{0,\dots,2^k-1\}$.

Perhaps this is the proof you already know.  It certainly is a follow-your-nose proof.