If the mean of a random process is constant, does it imply the process is first order stationary?
In the usual sense of the term, first-order stationarity means that the first-order distribution of all the random variables is the same: each $X_t$ has the same CDF, and so the same pdf (or pmf) too if the random variables are continuous (or discrete). If the random variables have a mean, then they all have the same mean. But, a sequence of independent
identically distributed Cauchy random variables is a first-order stationary process (as well as a strictly stationary process for that matter) but it cannot be said to have constant mean because none of the variables have a mean at all.

Since your response to my query on the problem statement is that your definition of first-order stationarity is the  standard one, then the answer to the question in the title of your question, viz.


  If the mean of a random process is constant, does it imply that the
  process is first-order stationary?


is NO, and an example of a process that has constant mean but is not
first-order stationary is a sequence of constant-mean random variables whose variances depend on time.

The answer to the question in the text, viz.


  Is a WSS process necessarily first-order stationary?


ia also NO.  Consider the random process
$\{X(t)\colon X(t)= \cos (t + \Theta), -\infty &lt; t &lt; \infty\}$
where $\Theta$ takes on four equally likely values $0, \pi/2, \pi$ and $3\pi/2$ and so $X(t)$ takes on four equally likely values
$\cos(t)$, $\cos(t+\frac{\pi}{2}) = -\sin(t)$, 
$\cos(t+\pi) = -\cos(t)$, and
$\cos(t+\frac{3\pi}{2}) = \sin(t)$.
  It is easy to see that in general $X(t)$ and $X(s)$ have different distributions and so the process is not first-order stationary. On the other hand, $E[X(t)] = 0$ for all $t$ while
\begin{align}
E[X(t)X(s)]&amp;= \left.\left.\frac 14\right[2\cos(t)\cos(s) + 2\cos(t+\pi/2)\cos(s+\pi/2)\right]\\
&amp;= \left.\left.\frac 12\right[\cos(t)\cos(s) + \sin(t)\sin(s)\right]\\
&amp;= \frac 12 \cos(t-s)
\end{align}
and so the process is wide sense stationary, but its first-order distribution does depend the choice of $t$ and so the process is not
first-order stationary.



Note: the above proof uses a very commonly accepted definition of wide sense stationarity as found in numerous books, online resources, Wikipedia etc, viz. a random process $\{X(t)\}$ is called a wide sense stationary (WSS) process if


$E[X(t)]$ is a constant, and
$R_{X}(t,s) = E[X(t)X(s)]$ depends only on the difference $t-s$ of the two arguments and not on the individual arguments $t$ and $s$.