Let us suppose we have two identical electrical bicycles A and B with identical batteries, both with some stored electrical energy $E$.
A is in motion with Kinetic energy $K_A$. It gets $E$ from its battery, and undergoes a speed increase of $V_A$ measured by its speedometer.
I will write the initial kinetic energy $K_A$ as
$$
K_A = \frac{1}{2}mv_1^2
$$
To simplify the analysis, I will ignore friction and air resistance, and will also assume that all the battery energy $E$ is transformed into kinetic energy:
$$
K_A + E = \frac{1}{2}m(v_1 + v_A)^2\;.\tag{1}
$$
B too is in motion, but with Kinetic energy $K_B$ > $K_A$. It gets $E$ from its battery, and undergoes a speed increase of $V_B$ measured by its speedometer.
The bikes are identical, so their mass is identical. So, I will write $K_B$ as
$$
K_B = \frac{1}{2}mv_2^2\;,
$$
where $v_2>v_1$ since we have stated that $K_B>K_A$.
Again, I assume all the battery energy is transformed into kinetic energy, so we have
$$
K_B + E = \frac{1}{2}m(v_2 + v_B)^2\;. \tag{2}
$$
Is it correct that $V_B < V_A$?
Yes. Solving Eq. (1) and Eq. (2) for $v_A$ and $v_B$, respectively, gives:
$$
v_A=\sqrt{v_1^2 + \frac{2E}{m}} - v_1\tag{3}\;,
$$
and
$$
v_B=\sqrt{v_2^2 + \frac{2E}{m}} - v_2\tag{4}\;.
$$
Or, rearranging, we can see that
$$
v_A^2 + 2 v_Av_1 = v_B^2+2v_Bv_2
$$
and so
$$
v_A^2 + 2 v_Av_1 > v_B^2+2v_Bv_1\;,\tag{5}
$$
since $v_2 > v_1$.
Without loss of generality, I will assume $v_A$, $v_B$, $v_1$, and $v_2$ are all positive (i.e., the bikes are traveling in the positive direction). Thus, rearranging Inequality (5) we have:
$$
v_A^2 - v_B^2 + 2v_1(v_A - v_B)>0
$$
or
$$
(v_A - v_B)\left(v_A + v_B + 2v_1\right) > 0
$$
or
$$
v_A - v_B >0\;.
$$
In other words, their speeds do not increase by the same amount?
Yes.
If so, by measuring the increase in speed, one can deduce the initial speed.
Yes. The initial speed can be determined from the known energy $E$, mass $m$, and speed increase $v_A$ as
$$
v_1 = \frac{E}{mv_A} - \frac{v_A}{2}\;,
$$
and similarly for $v_2$ with $v_A$ replaced by $v_B$.
This contradicts the Galilean principle of relativity.
No, it doesn't.
