Since $\varphi$ is a Möbius transformation it sends circles onto circles. The Orientation Principle (see Chapter III, 3.21 p.53 on John Conway: Functions of one complex variable I. Second edition) states that if $\varphi$ maps the circle $C_1$ onto the circle $C_2$ then everything on one side of $C_2$ must be send exclusively to only one side of $C_2$ (the sides depend on the orientation given).
In this case, we have that the unit circle $T$ is sent onto the imaginary axis (which is a circle on the Riemann Sphere), thus the imaginary axis is the boundary of $\varphi(T)$, that is everything on the inside of $T$ must be send only to the left side or only to the right side of the imaginary axis.
To check what side (that is to check what orientation is taken), we only need to check where $\varphi$ sends any point inside $T$. Then since a point inside the unit circle $T$, in this case $0$ is sent to a point on the right half plane $\varphi(0)=1$, we get that $\varphi(T)$ must be the right half plane.
As a corollary we get that $\varphi(\mathbb{C} \setminus T)$ is the left half plane, and you an check it for example by computing $\varphi(2)=-3$
Rouché's theorem:
Picking $z^4 + 2$ as the approximating function works well in fact. The fourth roots of $-2$ are
$$\sqrt[4]{2}\cdot \exp \biggl(\frac{\pi}{4}i + k\frac{\pi}{2}i\biggr)\,,\quad k \in \{0,1,2,3\}\,.$$
These lie on the diagonals, not on the imaginary axis, so we don't have to deal with zeros on the boundary of the half-disk $D_R = \{ z : \lvert z\rvert < R,\, \operatorname{Re} z > 0\}$ for large enough $R$. The zeros corresponding to $k = 0$ and $k = 3$ lie in the right half-plane, the other two in the left.
We have a problem if we only know Rouché's theorem in the form that $f$ and $g$ have the same number of zeros in $V$ if $\lvert f(z) - g(z)\rvert < \lvert g(z)\rvert$ for all $z \in \partial V$, however, since with the given $f$ and $g(z) = z^4 + 2$ this inequality doesn't hold on the whole boundary of $D_R$ (for large enough $R$; it holds for small $R$, but that isn't enough). For $z = i$ we have $\lvert g(i)\rvert = 3$ and
$$\lvert f(i) - g(i)\rvert = \lvert -2 -(5+\sqrt{2})i\rvert > 5 + \sqrt{2} > 3$$
(and furthermore $\lvert f(i) - g(i)\rvert > \lvert f(i)\rvert$, so swapping the roles of $f$ and $g$ doesn't help).
We get around this problem if we take a look at a standard proof of Rouché's theorem. In this proof, the conclusion is obtained from the observation that
$$N(\lambda) := \frac{1}{2\pi i} \int_{\partial V} \frac{g'(z) + \lambda\bigl(f'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(f(z) - g(z)\bigr)}\,dz$$
is constant on $[0,1]$. The premise $\lvert f(z) - g(z)\rvert < \lvert g(z)\rvert$ on $\partial V$ ensures that the denominator of the integrand never vanishes for $\lambda \in [0,1]$ and $z \in \partial V$. The integrand is jointly continuous in $z$ and $\lambda$, $\partial V$ has finite length, so $N$ depends continuously on $\lambda$. But $N$ is integer-valued, hence it must be constant.
So what is needed in the proof is not $\lvert f(z) - g(z)\rvert < \lvert g(z)\rvert$ for $z \in \partial V$, we only need the consequence that
$$f_{\lambda}(z) := g(z) + \lambda\bigl(f(z) - g(z)\bigr) = (1-\lambda)g(z) +\lambda f(z)$$
has no zeros on $\partial V$, for every $\lambda \in [0,1]$. This means that for every $z \in \partial V$ the straight line segment connecting $g(z)$ and $f(z)$ does not pass through $0$, or equivalently, that
$$\lvert f(z) - g(z)\rvert < \lvert f(z)\rvert + \lvert g(z)\rvert$$
for all $z \in \partial V$.
Now we have a form of Rouché's theorem whose premise is as symmetric in $f$ and $g$ as its conclusion, and the theorem is stronger since we weakened the hypothesis.
This version works for $g(z) = z^4 + 2$. On the semicircle $\lvert z\rvert = R,\, \operatorname{Re} z \geqslant 0$ we have
$$\lvert f(z) - g(z)\rvert < \lvert g(z)\rvert \leqslant \lvert f(z)\rvert + \lvert g(z)\rvert$$
for large $R$, and on the imaginary axis we have
$$\operatorname{Re} f_{\lambda}(z) = z^4 + 2\lambda z^2 + 2 = (z^2 + \lambda)^2 + (2 - \lambda^2) \geqslant 1\,,$$
so no $f_{\lambda}$ has a zero on the imaginary axis. (Note that it is frequently the case that different methods of showing $\lvert f(z) - g(z)\rvert < \lvert f(z)\rvert + \lvert g(z)\rvert$ are more convenient on different parts of the boundary. Here it is easy to show that he connecting line segments don't pass through $0$ for purely imaginary $z$, while estimating the moduli is cumbersome, and on the semicircle the estimate is straightforward.)
Argument principle:
A comment suggested the argument principle, and you replied
I did consider that, but I was stuck figuring out the integration.
The nice thing about the argument principle is that
$$\frac{1}{2\pi i}\int_{\partial V} \frac{f'(z)}{f(z)}\,dz$$
is always an integer, and thus it suffices to come close enough to know which one it is. This can of course still be quite difficult, but for the case of polynomials and their zeros in some half-plane it's not so bad (at least if the degree is low enough).
If we have a polynomial $p(z) = z^d + \dotsc$ of degree $d$ (we can without loss of generality assume the polynomial is monic), then for large $\lvert z\rvert$ we have
$$\frac{p'(z)}{p(z)} = \frac{d}{z} + O\biggl(\frac{1}{z^2}\biggr)\,.$$
Thus for sufficiently large $R$ the integral over the semicircle $S_R$ is
$$\int_{S_R} \frac{p'(z)}{p(z)}\,dz = \int_{S_R} \frac{d}{z} + O\biggl(\frac{1}{z^2}\biggr)\,dz = d\pi i + O\biggl(\frac{1}{R}\biggr)\,.$$
It remains to determine the change of argument along the diameter closing the contour accurately enough to get the final result. For this one can split the polynomial into parts that are real and purely imaginary respectively on the diameter. Using a parametrisation $t \mapsto a + tb$ of the diameter, one gets real polynomials $u,v$ such that $p(a + tb) = u(t) + iv(t)$, and it suffices to approximately locate the zeros of $u$ (or $v$) and determine the sign of $v(t)$ (resp. $u(t)$) and orientation of $t \mapsto u(t) + iv(t)$ to count how often $p(z)$ winds around $0$ on the diameter.
If $t_1 < t_2$ are two consecutive zeros of $u$, then we have $\pm$ half a winding around $0$ between $t_1$ and $t_2$ if $v(t_1)$ and $v(t_2)$ have different sign, and no winding if the signs are the same. We get $+\frac{1}{2}$ windings if $v(t_1) > 0 > v(t_2)$ and $u(t) < 0$ for $t_1 < t < t_2$ or $v(t_1) < 0 < v(t_2)$ and $u(t) > 0$ for $t_1 < t < t_2$, and we get $-\frac{1}{2}$ windings for the same $v(t_k)$ combinations with $u$ having opposite sign between $t_1$ and $t_2$. The (approximate) change of argument between the smallest/largest zero of $u$ and the endpoints ($E$) of the diameter is easy to determine from the sign of $v$ at the zero and the location of $p(E)$. For polynomials of large degree, this can be quite tedious, but since approximations to the zeros of $u$ suffice - and we need only consider pairs of zeros of $u$ between which $v$ changes sign - it's not too bad. And if we're lucky, one of $u$ and $v$ has constant sign on the whole diameter.
In our case, things work out particularly well. The coefficients of $f$ are real, the diameter lies on the imaginary axis since we're interested in the zeros in the right half-plane, so we have a simple split into the real and imaginary parts given by the parity of the exponent of $z$. On the imaginary axis $z^4 + 2z^2 + 2$ is real, and $\sqrt{2}z^3 - 5z$ purely imaginary. Writing $z = ti$ with $t \in \mathbb{R}$ on the imaginary axis, we have
$$z^4 + 2z^2 + 2 = (z^2+1)^2 + 1 = (1 - t^2)^2 + 1 \geqslant 1\,.$$
Thus the image of the diameter connects $f(iR)$ and $f(-iR)$ without leaving the right half-plane, and the change of argument along the diameter is just the difference of arguments between $f(-iR)$ and $f(iR)$. Since
$$f(\pm iR) = R^4 \mp\sqrt{2}R^3i + O(R^2) = R^4\biggl(1 \mp \frac{\sqrt{2}}{R}i + O(R^{-2})\biggr)$$
the arguments are $\sim \mp \frac{\sqrt{2}}{R}$, so the difference is $O(R^{-1})$ and
$$\int_{\partial D_R} \frac{f'(z)}{f(z)}\,dz = 4\pi i + O\biggl(\frac{1}{R}\biggr)\,.$$
We see that for large enough $R$ the total change of argument along the boundary of the half-disk is $4\pi i$, whence there are two zeros in the right half-plane (counting multiplicities, but the zeros are in fact simple).
Best Answer
Choose $g(z) = z^5 + 1$. The zeros of $g$ are $e^{k\pi i/5},\; k \in \{1,3,5,7,9\}$, two of which lie in the right half plane.
Then a slightly generalised version of Rouché's theorem tells you that $h$ also has two zeros in the right half plane.
The standard formulation of Rouché's theorem demands that $\lvert h-g\rvert < \lvert h\rvert$ (or $\lvert h-g\rvert < \lvert g\rvert$) on the boundary $\partial V$ of the region, but looking at the proof, whose main part is the observation that the number of zeros of $f_\lambda = g + \lambda(h-g)$ in $V$,
$$N_\lambda = \frac{1}{2\pi i}\int_{\partial V} \frac{g'(z) + \lambda\bigl(h'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(h(z) - g(z)\bigr)}\, dz$$
is independent of $\lambda \in [0,\,1]$, since no $f_\lambda$ has a zero on $\partial V$.
The role of the condition $\lvert h-g\rvert < \lvert g\rvert (\text{ or } \lvert h\rvert)$ on $\partial V$ is solely to guarantee that $g + \lambda(h-g)$ also has no zeros on $\partial V$. If we can determine that in a different way, the conclusion still holds.
For a large enough semicircle in the right half plane, we have $\lvert 5z^3 + 2z^2 + 4z\rvert < \lvert z^5 + 1\rvert$ since the RHS is a polynomial of higher degree than the LHS.
On the imaginary axis, although $\lvert g-h\rvert > \lvert g\rvert$ on some parts, expanding shows
$$g(it) + \lambda\bigl(h(it) - g(it)\bigr) = 1 + it^5 + \lambda(-5it^3 - 2t^2 + 4it) = (1-2\lambda t^2) + it(t^4 - 5\lambda t^2 + 4\lambda),$$
and the real part vanishes only for $\lambda = \dfrac{1}{2t^2}$ (with $t^2 \geqslant \frac12$, since $\lambda \leqslant 1$), when the imaginary part is
$$t\left(t^4 - \frac{5}{2} + \frac{2}{t^2}\right) = \frac1t\left(t^6 - \frac{5}{2}t^2 + 2\right),$$
and the polynomial $x^3 - \frac52x + 2$ has no positive real roots, so we conclude $g(z) + \lambda(h(z)-g(z)) \neq 0$ on the imaginary axis.