PART-I: I would like to provide the general proof considering the title of the question and imposing no specific properties on the signal $x(t)$ other than having a CTFS representation.
The following is a simple analysis to conclude that the CTFS coefficients of any signal $x(t)$ and that of the DC removed signal are equivalent. (except $a_0$ of course).
Consider a continuous-time periodic signal $x(t)$ with period $T$ divided into two components: $x_{dc}$ and $x_{ac}$, with periods $T$ also, where $x_{dc}$ is the pure DC component of $x(t)$ and $x_{ac}$ is the pure AC component of $x(t)$, then we have:
$$ x(t) = x_{dc}(t) + x_{ac}(t)$$
Computing the CTFS coefficient $a_k$ of $x(t)$ yields:
$$
\begin{align}
a_k &= \frac{1}{T} \int_{<T>} (x_{dc} + x_{ac}) e^{-j k \frac{2\pi}{T} t } dt \\
&= \frac{1}{T} \int_{<T>} x_{dc} e^{-j k \frac{2\pi}{T} t } + \frac{1}{T} \int_{<T>} x_{ac} e^{-j k \frac{2\pi}{T} t } \\
a_k &= b_k + c_k \\
\end{align}
$$
where $b_k$ and the $c_k$ are the CTFS coefficients of DC and AC parts of $x(t)$.
By definition of any DC signal, it's known that $b_k = 0$ for all $k \neq 0$ and by defition of any AC signal it's known that $c_0 = 0$. Then using the relation $a_k = b_k + c_k$ we get the following:
$$a_0 = b_0 + c_0 = b_0$$
and
$$a_k = 0 + c_k = c_k ~~~,~~~ \text{ for all } k \neq 0 $$
From which we define :
$$
b_k = \begin{cases} a_0 ~~~&, ~~~\text{ for } k = 0 \\ 0 ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}
$$
and
$$
c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases}
$$
Hence we conclude that the CTFS coefficients, $a_k$, of any periodic signal $x(t)$ and the CTFS coefficients $c_k$ of DC-removed part, $x_{ac}$, are the same for all $k \neq 0$.
PART-II: Based on OP comments, the relation for an even and real signal, is the following.
For a signal $x(t)$ which is real and even we have
$$x(t) = x(t)^{*} = x(-t) = x(-t)^{*}$$
and the associated CTFS coefficients has the property of
$$ a_k = a_{-k}^{*} = a_{-k} = a_k^{*} $$
which indicates that the coefficients $a_k$ are also real and even.
Using this, we can obtain the trigonometric (cosine) Fourier series coefficients as.
$$
\begin{align}
a_k &= \frac{1}{T} \int_{<T>} x(t) e^{-j k \frac{2\pi}{T} t } dt \\
&= \frac{1}{T} \int_{<T>} x(t) \left( \cos( k \frac{2\pi}{T} t) + j \sin(k \frac{2\pi}{T} t) \right) dt \\
&= \frac{1}{T} \int_{<T>} x(t) \cos( k \frac{2\pi}{T} t) dt + j \frac{1}{T} \int_{<T>} x(t) \sin( k \frac{2\pi}{T} t) dt\\
a_k &= \mathcal{Re}\{a_k\} + j ~~ \mathcal{Im}\{a_k\} \\
\end{align}
$$
Now since the property states tat $a_k$ are real, then the imaginary part is zero and we have:
$$
a_k= \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt
$$
Furtermore since $x(t)$ is also even; $x(t)=x(-t)$, then we also have
$$
\boxed{ a_k= \frac{2}{T} \int_{0}^{T/2} x(t) \cos( k \frac{2\pi}{T} t) dt }
$$
As the trigonometric cosine series coefficients of the real and even signals.
In addition to this, for a real & even signal, $x(t)$ of period $T$, which has no DC part, the following is also observed:
$$ x(t-\frac{T}{2}) = -x(t) $$
And based on the time-shift property of CTFS we can conclude that
$$ a_k ~~e^{-j\frac{2\pi}{T}k \frac{T}{2} } = - a_k $$
$$ a_k ~ e^{-j\pi k } = - a_k \implies a_k = \begin{cases} -a_k &, k=0,\pm 2, \pm4,... \\ a_k &, k=\pm 1,\pm 3,...\\ \end{cases}$$
Which indicates that the CTFS coefficients $a_k = 0$ for $k=2m$ (even) for a real, even, (and having no DC) signal $x(t)$. Indeed we can get rid of the DC removed adjective and state for all real & even signals, as the DC will only affect $a_0$ being non-zero.
PART-III: Finally apply these to the example signal to see that it works.
The signal defined as:
$$
x(t) =
\begin{cases} t &, 0<t<1 \\ 2-t &,1<t<2 \\ \end{cases}
$$
Then for the CTFS coefficeints (in the trigonometric form) we have:
$$a_k = \frac{2}{T} \int_{0}^{1} t \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt $$
A simple by-parts integration yields the following result:
$$\begin{align}
a_k &= \int_{0}^{1} t \cos(k \pi t) dt \\
&= \frac{t ~\sin(\pi k t) }{\pi k} |_0^1 -\int_{0}^{1} \frac{\sin(k \pi t)}{\pi k} dt \\
&= \frac{\sin(\pi k) }{\pi k} + \frac{1}{\pi k} \left( \frac{ \cos(\pi k) - 1}{\pi k} \right)\\
\end{align}
$$
we conclude that
$$\boxed{ a_k = \frac{ \pi k ~\sin(\pi k) + \cos(\pi k) - 1}{ \pi^2 k^2 } }$$
Form which it can also be seen that
$$a_k = \begin{cases} 0.5 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases} $$
Note that the term $\sin(\pi k)$ can be ignored (as it's all zero) except for $k=0$. Also note that those zero values $a_k$ turns out to be the case without assuming half wave symmetry.
Finally, we shall compute the CTFS $c_k$ coefficeints of the DC removed signal $x_{ac}(t)$ to see if they are equivalent. From the definition of the signal we see that
$$
x_{ac}(t) = x(t) - 0.5 =
\begin{cases} t-0.5 &, 0<t<1 \\ 1.5-t &,1<t<2 \\ \end{cases}
$$
then the $c_k$ become:
$$c_k = \frac{2}{T} \int_{0}^{1} (t-0.5) \cos(\frac{2\pi}{T} k t) dt = \int_{0}^{1} t \cos(k \pi t) dt - \int_{0}^{1} 0.5 \cos(k \pi t) dt $$
Note that this integral is the same for the case of $a_k$. The only diference is in the last term which is $1$ for $k=0$ and $0$ for all $k\neq 0$ and can be ignored for $k \neq 0$. Then the equation relating to $c_k$ will be identical to that of $a_k$ except at $k=0$ which yields:
$$c_k = \begin{cases} 0 &, k=0 \\ \frac{-2}{\pi^2k^2} &, k=\pm 1, \pm 3,... \\ 0 &, k=\pm 2, \pm 4,...\\ \end{cases} $$
hence we again concluded that
$$
\boxed{ c_k = \begin{cases} 0 ~~~&, ~~~\text{ for } k = 0 \\ a_k ~~~&, ~~~\text{ for all } k \neq 0 \\ \end{cases} }
$$
