Consider a map $f\colon S^1 \to S^1 \vee S^1$ that maps upper half of a circle to the first summand in orientation preserving way, and lower half to the second circle in orientation reversing way. Also denote $i_1$ and $i_2$ inclusions of $S^1$ to $S^1 \vee S^1$ as two summands. Composition of $f$ with folding map $p\colon S^1\vee S^1 \to S^1$ (both summands in orientation-preserving way) is evidently homotopic to a constant map. Looking on $H_1$ we have $\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}\oplus \mathbb{Z} \xrightarrow{p_*} \mathbb{Z}$.
$p_*$ is addition: $S^1 \xrightarrow{i_1} S^1\vee S^1 \xrightarrow{p} S^1$ is identity (similary for $i_2$). $i_1$ induces $\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$, to make sure it is an inclusion of the first summand we may use naturality of exact sequences for a map $(S^1, *) \xrightarrow{i_1} (S^1\vee S^1, \mathrm{im}\, i_2)$. Now we know $p_*(1, 0) = 1$ and $p_*(0, 1) = 1$, by linearity $p_*$ is addition.
Thus $f_*$ maps $1\in H_1(S^1)\cong \mathbb{Z}$ to some pair $(x, -x)\in H_1(S^1\vee S^1)$. To find out what $x$ is, consider the composition $S^1 \xrightarrow{f} S^1 \vee S^1 \to S^1$ where the second map collapses second summand to a point (it induces projection on a second coordinate in $\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ too, e.g. by naturality of exact sequences of pairs). By construction this map is homotopic to identity. Similary for the second summand in $S^1\vee S^1$ we get that orientation-reversing map induces $-1$ on homology.
That's quite cumbersome to read, but actually very straightforward.
UPDATE
As mentioned in comments, here is proof reversing orientation induces identity on $H_0$. $* \to S^1$ and $*\to S^1 \to S^1$ (the second map is reversal of orientation) are equal as maps from a point, thus induce the same map on $H_0$. If we know that the inclusion of a base point in $S^1$ induces non-zero map on $H_0$, we're done, since automorphism of $H_0(S^1) \cong \mathbb{Z}$ fixing a non-zero element must be identity. To see this, use the naturality of long exact sequences of pairs for a map $(S^1, *) \to (*, *)$. The first commutative square (with non-relative $H_0$s) consists of three identity mapping of a point and a map induced from inclusion of a base point, and cannot commute unless $H_0(*) \to H_0(S^1)$ is an inclusion.
(Reply to a comment: nullhomotopic map induces zero on $H_1$ since it factors as map $X \to * \to Y$)
Also it seems to me that $X \hookrightarrow X \vee Y$ induces inclusion to the first summand in homology not due to functoriality properties of long exact sequences but this is just the way we identify $H_i(X\vee Y)$ with $H_i(X)\oplus H_i(Y)$. That is, we look at the pair $(X \vee Y, Y)$, write long exact sequence, then apply excision to get rid of relative homology and get $H_i(Y)$ instead. After that, we note that a lot of arrows in long exact sequence admit splitting and define the isomorphism $H_i(X \vee Y) \to H_i(X) \oplus H_i(Y)$ the way inclusion of $X$ in a wedge sum is inclusion on first summand. Thus some arguments by naturality above are not needed.
The augmented chain complex of a space $X$ is
$$...\rightarrow S_2(X)\stackrel{\partial}{\rightarrow} S_1(X) \stackrel{\partial}{\rightarrow} S_0(X) \stackrel{\epsilon}{\rightarrow} \mathbb Z \to 0 .$$
This is defined also for $X = \emptyset$, but in fact is usually only considered for non-empty $X$. The reduced homology groups of $X$ are the homology groups of the augmented chain complex, therefore we have $\tilde H_n(X) = H_n(X)$ for $n > 0$. For $n = 0$ we get $\tilde H_0(X) = \ker(\epsilon)/\text{im}(\partial)$ which can be identified with a subgroup of $H_0(X) = S_0(X)/\text{im}(\partial)$. Moreover, one can easily show that $\tilde H_0(X) \approx \ker(p_* : H_0(X) \to H_0(*))$, where $p : X \to *$ is the unique map to a one-point space $*$. Note that $\tilde H_0(X) = 0$ for $X = \emptyset$.
What happens for $n = -1$? If $X \ne \emptyset$, then $\epsilon$ is surjective and $\tilde H_{-1}(X) = 0$, but if $X = \emptyset$, then we get $S_0(X) = 0$ and $\tilde H_{-1}(X) = \mathbb Z$. This is why Hatcher says that we should choose $X$ to be non-empty to avoid getting nonzero reduced homology groups of negative degree. However, it is no real problem to allow also $X = \emptyset$.
Without assuming $A \ne \emptyset$, the exact sequence of $(X,A)$
$$…\rightarrow \tilde H_n(A)\rightarrow \tilde H_n(X) \rightarrow H_n(X,A)\rightarrow …$$
ends with
$$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \tilde H_{-1}(A) \rightarrow \tilde H_{-1}(X) \to 0$$
You see that for $A \ne \emptyset$ we get $\tilde H_{-1}(A) = \tilde H_{-1}(X) = 0$ which yields Rotman's and Hatcher's sequence. For $A = \emptyset$ we get
$$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \mathbb Z \rightarrow \tilde H_{-1}(X) \to 0$$
where $\tilde H_{-1}(X) = 0$ if $X \ne \emptyset$ and $\tilde H_{-1}(X) = \mathbb Z$ if $X = \emptyset$.
You see that we do not get contradictions.
Best Answer
No.
The "classical" Eilenberg-Steenrod axioms are homotopy invariance, exactness, excision and dimension. These describe ordinary homology theories. It is well-known that for finite CW-pairs the homology groups $H_n(X,A)$ are (up to natural isomorphism) uniquely determined by the coefficient group $G = H_0(*)$, where $*$ is a one-point space. In fact, they agree with the singular homology groups of $(X,A)$ with coefficients in $G$. In particular $H_n(X,A) = 0$ for $n < 0$.
Beyond finite CW-pairs things are more sophisticated. In
one can find examples of non-trivial ordinary homology theories with zero coefficient group. There are infinite CW-complexes $X$ with nonvanishing homology groups. Making a dimension shift ($H'_n = H_{n+k}$ for some $k \in \mathbb N$) we can achieve that $H'_n(X) \ne 0$ for negative $n$. The theory $H'_*$ has coefficient group zero, if you do not like that consider $H'_* \oplus H^{sing}_*$, where $H^{sing}_*$ is singular homology with $\mathbb Z$-coefficients.