Let's assume we have an object of mass $m$ with two forces acting: gravity vertically down with magnitude $F_g = mg$, and a constant applied horizontal force (say, to the left, as in the OP's picture) with the same magnitude, $F_{\rm app} = mg$. Just looking at the free-body diagram, we see that the net force, and thus the acceleration, must point down and to the left. And because the forces are both constant, the acceleration will also be constant.
Let's choose an $x$-$y$ coordinate system with $x$ pointing left and $y$ pointing down. Then Newton's Second Law gives:
\begin{align}
\vec{F}_{\rm net} &= m \vec{a}\\
\vec{F}_g + \vec{F}_{\rm app} &= m \vec{a}\\
x: \quad F_{g,x} + F_{{\rm app},x} &= m a_x \\
0 + F_{\rm app} &= m a_x \\
m g &= m a_x \quad \rightarrow \quad a_x = g\\
y: \quad F_{g,y} + F_{app,y} &= m a_y \\
F_g + 0 &= m a_y \\
m g &= m a_y \quad \rightarrow \quad a_y = g
\end{align}
So the constant acceleration vector is $\vec{a} = (a_x, a_y) = (g, g)$.
Now if acceleration is constant, the equations for motion with constant acceleration can be used. In the $x$ direction we have:
\begin{align}
x(t) &= x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \\
&= 0 + 0 + \frac{1}{2} g t^2
\end{align}
and
\begin{align}
y(t) &= y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \\
&= 0 + 0 + \frac{1}{2} g t^2
\end{align}
where I assumed that the particle starts at the origin with zero initial velocity, $\vec{v}_0 = (v_{0x}, v_{0y})= (0,0)$.
The trajectory is the path we see it take through space, i.e., the function $y(x)$. So we must eliminate time from the above two equations:
$$
x(t) = \frac{1}{2} g t^2 \quad \rightarrow \quad y(t) = \frac{1}{2} g t^2 = x(t)
$$
or,
$$
y(x) = x
$$
The trajectory is a straight line down and to the left with slope 1.
Note that if you wanted to find the time for the object to hit the ground, you would just use the $y$ equation:
$$
y(t) = \frac{1}{2} g t^2 \quad \rightarrow \quad H = \frac{1}{2} g t_{\rm hit}^2 \quad \rightarrow \quad t_{\rm hit} = \sqrt{\frac{2H}{g}}
$$
which is exactly what you would find for an object dropped from the same height. So, no, the horizontal force does not change the time of fall.